# WORK, ENERGY, AND POWER CHAPTER 7 by add37610

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```									CHAPTER 7                      WORK, ENERGY, AND POWER

How much work do you do on the barbell during
ActivPhysics can help with these problems:
this time? (c) You lower the barbell to the ground.
Activity 5.1                                                 Now how much work do you do on it?
Section 7-1: Work
Solution
Problem                                                    (a) If we assume the barbell is lifted at a constant
1. How much work do you do as you exert a 75-N             velocity by a vertical applied force equal to the weight
force to push a shopping cart through a 12-m-long       and positive upward parallel to the displacement, then
supermarket aisle?                                      Wapp == mg ~y == (45 kg)(9.8 m/s2)(2.5 m) == 1.10 kJ.
(b) 'Just holding the weight stationary, you must still
Solution                                                   exert an applied force of my to balance gravity, but
If the force is constant and parallel to the displace-     the displacement through which the force acts is zero.
ment, W=F.~r=F~r==(75N)(12m)                 == 900J.      Hence the work done on the barbell is also zero.
(Actually, individual muscle fibers are continually
Problem                                                    contracting even though the overall muscles are
2. If the coefficientof kinetic friction is 0.21, how      stationary, so internal work is being done in your
much work do you do when you slide a 50-kg box at        muscles and you feel tired just holding a weight.)
constant speed across a 4.8-m-wide room?                 (c) When the weight is lowered at constant velocity,
the upward applied force, mg, is opposite to the
Solution                                                    displacement downward, ~y == -2.5 m; hence the work
If you push parallel to a level floor, the applied force    done on the barbell is negative, Wapp == -1.10 kJ.
equals the frictional force (since the acceleration is
zero), and the normal force equals the weight. The         Problem
applied force is constant and parallel to the              5. The world's highest waterfall, theCherun-Meru in
displacement, so Wa == Fa ~r == !k ~r == ItkN ~r ==           Venezuela, has a total drop of 980 m. How much
2
Itkmg ~r == 0.21(50 kg)(9.8 m/s )(4.8 m) == 494 J.            work does gravity do on a cubic meter of water
dropping down the Cherun-Meru?
Problem
3. A crane lifts a 650-kg beam vertically upward 23 m,     Solution
then swings it eastward 18 m. How much work does        The force of gravity at the Earth's surface on a cubic
the crane do? Neglect friction, and assume the          meter of water is Fg == mg== 9.8 kN vertically
beam moves with constant speed?                         downward (see the inside book cover of the text for
the density of water). The displacement of the water
Solution                                                   is parallel to this, so the work done by gravity on the
Lifting the beam at constant speed, the crane exerts a     water is Wg == Fg ~y == (9.8 kN)(980 m) == 9.6 MJ.
constant force vertically upward and. equal in
magnitude to the weight of the beam. During the            Pr6blem
horizontal swing, the force is the same, but is            6. A meteorite plunges to Earth, embedding itself
perpendicular to the displacement. The work done is :         75cm in the ground. If it does 140 MJ of work in
F. ~r == (mgj) .(~yj + ~xi) == mg ~y == (650 kg) x ,          the process, what average force does the meteorite
(9.8 m/s2)(23 m) == 147 kJ.                                   exert on the ground?

Problem                                                    Solution
The average force exerted by the meteorite parallel to
4. You lift a 45-kg barbell from the ground to a height
of 2.5 m. (a) How much work do you do on the            its penetration into the ground is Fav == W/ ~x ==
barbell? (b) You hold the barbell aloft for 2.0 min.    140 MJ/O.75 m == 187 MN, or about 21,000 tons.
108    CHAPTER 7

Problem                                                     Problem
7. You slide a box of books at constant speed up a          9. A locomotive does 7.9 x 1011 J of work in pulling a
30° ramp, applying a force of 200 N directed up the.        3.4x105-kg train 180 km. What is the average force
slope. The coefficient of sliding friction is 0.18.         in the coupling between the locomotive and the rest
(a) How much work have you done when the box                of the train?
has risen 1 m vertically? (b) What is the mass of
the box?                                                 Solution
Solution                                                    IT we define the average force by W = Fav !1.r, then
Fav = 7.9x1011 J/180 km = 4.39 MN. (The train's
(a) The displacement up the ramp (parallel to the
mass is not required to answer this question.)
applied force) is !1r = 1 m/ sin 30° = 2 m, so Wa =
Fa' ~r= (200 N)(2 m) = 400 J. (b) We could easily
Problem
solve Newton's second law, with zero acceleration, to
find the mass, m = Fa/ g(sin (J + fJ.k cos (J), but it is   10. An elevator of mass m rises a distance h up a
instructive to obtain the same result using the concept         vertical shaft with upward acceleration equal to
of work. The work done by gravity is F 9 • ~r   =               one-tenth g. How much work does the elevator
cable do on the elevator?
Fg•(!1xi+ !1yj)= -mg !1y= -m(9.8 m/s2)(l m)             =
-m(9.8 J /kg) (see Example 7-9). The work done by
friction is fk • ~r = - fk!1r = -fJ.kN !1.r = -J1.kx
Solution
(mg cos 3()O)!1.r = -0.18m(9.8 m/s2)(2 m) cos 30° =         To give the elevator a constant upward acceleration
-m(3.06 .J/kg). The total work is zero (v is constant       all = 0.1 g, the tension in the cable must satisfy
in the work-energy theorem), so 0= 400 J - mx                          =             =
T ~ mg mall' or T m(g + all) = 1.1mg. Acting
(9.8 + 3.06) J/kg, of m  =   400 J/(12.86 Jjkg) =           over a parallel displacement !1y = h upward, the
31.1 kg.                                                    tension does work WT = T!1.y       =1.1 mgh on the
elevator.

Section 7-2: Work and the Scalar Product
Problem
11. Show that the scalar product obeys the
distributive law: A. (B   C)+ =A . B A . C.+
Problem 7 Solution.
Solution
Problem                                                     This follows easily from the definition of the scalar
8. Two people push a stalled car at its front doors,        product in terms of components: A. (B + C) =
each applying a 280-N force at 25° to the forward        Ax(Bx + Gx) + AII(BIi + Gil) + Az(Bz + Gz) = AxBx   +
direction, as shown in Fig. 7-25. How much work          AIiBIi   + AzBz + AxGx + AIiGIi + AzGz = A . B + A. C.
does each do in pushing the car 5.6 m?                      With more effort, it also follows from trigonometry.
First:

Dsin«(JD - (JB) = Gsin(ec      - (JB)   (law of sines),
D = JB2        + G2 + 2BGcos«(Jc   - (JB)   (law of cosines),

FIGURE 7-25 Problem 8.

Solution
From symmetry, the displacement is in the forward
- direction, so the work done by each person is Wa =
Fa.!1r = Fa !1r cos e = (280 N)(5.6 m) cos 25° =
1.42 kJ.                                                                         Problem 11 Solution.
CHAPTER 7      109

Problem
14. Given the following vectors:
A has magnitude 10 and points 30° above the :rraxis
together give                                                     B has magnitude 4.0 and points 10° to the left of the
D cos(9v - 9B) == B   + G cos(9c   - 9B)'                    y-axis
C = 5.6i - 3.1j
Second:                                                                   +
D = 1.9i 7.2j,
A.D=ADcos9v                                                       compute the scalar products (a) A. B (b) C - D
=AD[cos(9v      - 9B) COS9B - sin(9v - 9B) sin9B]             (c)B-C.                        .
=A[B + Gcos(9c -9B)]cos9B
. Solution
-A[Gsin(9c - 9B)] sin9B
(a) The angle between A and B is 60° + 10° = 70°, so
=ABcos9B +AG{cos(9c - 9B)cos9B                          A-B = (10)(4) cos 70° = 13.7 (Note: units are ignored
- sin(9c - 9B) sin 9B]                                 in this problem.) (b).In terms of components, C -D =
=A.B+A.C.                                               (5.6)(1.9) ~ (-3.1)(7.2) = -11.7. (c) Express B in
components (4 cos 1000ff.4 sin 1000j), or C by
We used the identity for cos (a + {3)from Appendix. A.        magnitude and direction (6.40 at 29.0° below the
:rraxis). In either case, B-C=(4cos1000)(5.6)  +
Problem                                                       (4 sin 100°)(-3,1) = 4(6.4) cos 129° = -16.1.
12. (a) Find the scalar products i. i, j. j and k. k. "
(b) Find the scalar products i. j, j. k, k. i.            Problem
(c) Use the distributive law to multiply out the          15. (a) Find the scalar product of the vectors ai + bj
scalar product of two arbitrary vectors                       and bi - aj, and (b) determine the angle between
A=Azi+Ayj+Azk           and B=Bzi+Byj+Bzk.                    them. (Here a and b are arbitrary conStants.)
Then use the results of parts (a) and (b) to verify
Equation 7-4.                                             Solution
(a) (ai+bj)-(bi-aj)=ab+b(-a)        =0. (b) The
Solution                                                      vectors are perpendicular, so 9 = 90°.
(a) The dot product of any vector with itself equals its
magnitude squared, A. A = A 2 cos 0° = A 2• For any           Problem
unit vector, fi.. fi.= 1. (b) If two vectors are
16. Rework Example 7-5 using Equation 7-3 instead of
perpendicular, their dot product is zero, A -B =
Equation 7-4.
AB cos90° = 0 if A.L B. The unit vectors i, j, and k
are mutually perpendicular; hence i- j = j - k = k -i = O.
(c) A-B=(Azi+Ayj+Azk)              - (Bzi + Byj + Bzk) =
Solution
F has magnitude v'4902 + 2302 N and direction 9 =
AzBzi. i + AzByi -j + AzBzi -k + AyBJ -i +
tan-1(230 490 = 25.1° above the x-axis; I~rl =
AyBy~' j_+AyBzj- k + AzBzk -j + AzByk -j +
1.7 + 1;90 mat 47.0° above the x-axis. Therefore,
=                 +
AzBzk -k AzBz,+ AyBy AzBz. (We also used the                  F -~r= (541 N)(2.60 m) cos(47.00 - 25.1°) = 1.30 kJ.
commutative law, i. j = j -i = 0, etc.)
Problem
Problem
17. Use Equations 7-3 and 7-4 to show that the angle
13. One vector has magnitude 15 units, and another
\ between the vectors A=azi+ayj,      and B=bzi+
6.5 units. Find their scalar product if the angle
byj is
between them is (a) 27° and (b) 78°.

Solution
(a) A-B=(15u)(6.5u)cos27°       = 86.9u2• (b) A.B=
(15u)(6.5u) cos 78° = 20.3u2• (Note: We used u as the          Solution
symbol for the unspecified units.)
A-B=azbz + ayby = Ja~ +a~Jb~ + b~cosB, from
which the desired expression for () follows directly.
110      CHAPTER 7

Problem                                                      Problem
18. Find the angles between all three pairs of the            21. A force F=67i+23.i+55k        N is applied to a
vectors A= 3i+2.i, B::: -i+6.i, C=7i-       2j.               body as it moves in a straight line from rl =
16i+31j to r2 =2li+     lOj+ 14k m. How much
Solution                                                          work is done by the force?
The result of the previous problem gives the angle
between any two vectors as 8AB = cos-l(A. B/AB).
Solution
(We are using 0AB for the magnitude of the angle             'iW=F.~r=F.        (r2 - rl) = [67(21-16)             + 23(10   -
between A and H, which by definition is in the range         :!31) + 55(14)] N.m = 622 J.
0° to 180°.) Therefore:
'Problem
O=cos-l           [.     3(-1)+2(6)           ] =6580     22. A rope pulls a box a horizontal distance of 23 m,
AB              v'32 +22/(-1)2        +62       .,        as shown in Fig. 7-26. If the rope tension is 120 N,
and if the rope does 2500 J of work on the box,
o         = cos-l [      3(7) + 2( -2) ] = 49.6° and          what angle does it make with the horizontal?
AC.               v'I3/72 + (-2)2           '
Solution
8      -     -1[(-1)7+6(-2)]_1150
Be - cos         v'37 v'53        -        .          W = 2500 J = (120 N)(23 m) cosO, so 0 = 25.1°.
(Note that for any three vectors in a plane, one of the
above angles is equal to the sum of the other two.)

Problem                                                                                                     23~=Ax
19. Find the work done by a force F = 1.8i + 2.2j N as
it acts on an object moving from the origin to the
point r = 56i + 31j m.                                               FIGURE 7-26 Problem 22 Solution.

Solution
Section 7-3:      A Varying Force
The force is constant, and r is the displacement from
the origin, so the work done by F is W = F • r =              Problem
[(1.8)(56) + (2.2)(31)] N . m = 169 J.                        23. Find the total work done by the force shown in
Fig. 7-27 as the object on which it acts moves
Problem                                                           (a) from x = 0 to x = 3 kill; (b) from x = 3 kill to
20. A force F = 14i + llj N acts on an object. Find               x=4km.
the work done by the force if the object moves
50
from the origin to the point (a) 28i + 22j m and
(b) 22i - 28j m.                                                            40
.......
~ 30
..•
Solution                                                                    ~20
The work done by a constant force F acting over a                              10
displacement r from the origin is W = F • r. For the
1        2       3      4
given vectors, the scalar product can be calculated                                         Distance (Ian)
from Equation 7-4, Wa = (14x28 + llx22) J =
634 J for part (a), or Wb = (14x22 -llx28)       J = 0 for                    FIGURE 7-27 Problem             23.
part (b). However, one could observe that F is parallel
to r in part (a), since the slopes are equal
(Fy/ Fx = y / x ), and that F is perpendicular to r in
Solution
part (b), since the product of the slopes is minus one        The work done by a one-dimensional force is
((Fy/Fx)(Y/x) =: -1). Then Equation 7-3 gives                 (Equation 7-8) W =      1:;
F(x) dx. From Fig. 7-27,
Wa = 2(142 + 112) J and Wb = O.                               F(x) is a linear function in the two intervals specified:
(40 N/3 km)x,                 for 0 ::; x ::;3 km.
F(x) =      40 N - (40 N/km) x
{
(x - 3 km),                  for 3 km ::; x ::;4 km.
CHAPTER7         111

(Use the slope/int~rcept. equation for a straight line,    Solution
y = mx  + b, to verify this.) Therefore:                   (a) W =      1:: F dx   = fgmax2dx       =   la   Ix31~m =
(a)   WO-3    =   1 G~~)
3km
x dx
2
l(5 N/m )(6 m)3 = 360 J. (b) W::::l E~=l F(Xi) !::J.Xi,
where Xi = 1 m, 3 m, 5 m are the midpoints, and
= (40 N)     (3 km)2 = 60 kJ                 !::J.Xi= 2 m. Then W::::l (5 N/m2)(12 + 32 + 52) x
=
h:~(4~:)(4
3.krn        2           '               m2(2 m)       350 J. The percent error is only 8 =
100(360 - 350)/360 = 2.78%. (c) Now, Xi = 0.5, 1.5,
(b)   W3-~ =                          km - x)dx            2.5,3.5, 4.5, and 5.5 (in meters), and !::J.Xi 1 m.       =
W ::::l E~=l F(Xi) !::J.Xi= (5 N/m2)(0.5 m)2(12 + 32 +
(40 N) I(4 krn)x-
4km
=    -                       ~ 21            52 + 72 + 92 + 112)(1 m) = 357.5 J, and 8 = 100(2.5)..;-
2
krn                     2 3km          360 = 0.694%. (d) W::::l E~:l F(Xi)!::J.Xi            =
(5 N/m )x
=   20 kJ.                                   (0.25 m)2(1  2 +32 + ... + 232)(0.5 m)            =
359.375 J
(Of course, the triangular areas under the force vs        with 8 = 0.174%. (The direct calculation of the sum is
distance curve could have been calculated in one's         tedious, but we can use the formula for the sum of the
head; however, it is instructive to understand the         squares of the first n numbers, namely E~ k2 =
general method for evaluating Equation 7-8.)               ~n(n + l)(n + 2). The sum in question is Ei:l x
(2k -1)2 = E~:l         k2   -   Ei~1(2k)2   = 4324 - 2024 =
Problem                                                    2300.)
24. Findthe total work done by the force shown in.
Fig. 7-28, as the object on which it aCts moves :      Problem
from x = 0 to x = 5.0 m.                               26. A spring has spring constant k = 200 N/m. How
much work does it take to stretch the spring
Solution                                                       (a) 10 em from equilibrium and (b) from 10 cm to
The work is the area under the graph between x = 0             20 em from equilibrium?
and 5 m, as shown. The seven squares and four
triangles represent 10.5 J of work.                        Solution
(a) W = ~kx2 (from Equation 7-10) = ~(200 N/m) x
(0.1 m)2 = 1 J. (b) W            =
~k(x~ - x~) = ~(200 N/m)x
[(0.2 m)2- (0.1 m)2] = 3 J. (The work to stretch the
spring from 9 to 20 em is four times the work in
part (a), or 4 J, so the work in part (b) is
4 J -1 J
\
= 3 J.)

Problem
2     3     4    5      6       27. A certain amount of work is required to stretch
Distance (m)
spring A a certain distance. Twice as much work
FIGURE 7-28 Problem 24 Solution.
is required to stretch spring B half that distance.
Compare the spring constants of the two springs.

Problem                                                    Solution
25. A force F acts in the x direction, its magnitude       We are given     WA = !kAx2                       !k
and WB = B(x/2)2         =
given by F = ax2, where x is in meters, and a is       !kBX2    = 2WA = kAX2•           Therefore kB = 8kA•              .
. 2
exactly 5 N/m . (a) Find an exact value for the
work done by this force as it acts on a particle       Problem
moving from x = 0 to x      =  6 m. Now find           28.. On graph paper, draw an accurate force-distance
approximate values for the .work by dividing the             curve for the force of Example 7-8, and obtain a
area under the force curve into rectangles of width        ..solution to that example by determining
!
(b) !::J.x= 2 m; (c) !::J.x= 1 m; (d) !::J.x= m with         graphically the area under the curve.
height equal to the magnitude of the force in the
center of the interval. Calculate the percent error    Solution
in each c.ase.                                         We can superimpose a grid on Fig. 7-16 and count
rectangles. The area represented by one rectangle is
112    CHAPTER 7

1 kJ, and there are about six and one half such                      Solution
rectangles under the curve between x = 0 and 10 m.                   The work is the area under a cosine curve, which the
Thus, W ~ 6.5 kJ.                                                    student may plot if so required. The force reverses
direction when x = !1rXo ~ 20.4 m. The "area"
1500
between this and 37 m lies below the 3raxis and must
be counted as negative.

Problem
31. A force given by F = a..jX acts in the x direction,
where a = 9.5 N/ml/2. Calculate the work done
2       4     6      B    10                   by this force acting on an object as it moves
Position, % (Ill)                        (a) from x = 0 to x = 3 m; (b) from x = 3 m to
x = 6 m; (c) from 6 m to 9 m.
FIGURE    7-16 Problem 28 Solution.
Solution
Problem                                                              From Equations 7-8 and 7-9:
29. A force F. acts in the x direction; its x component                                      _lx2          1/2dx _ ax
32
/ 1
x2
W
is given by F = Fo cos(x/xo), where Fo = 51 N                                   Xl-+X2   -   Xl   ax         - (3/2)        Xl

and Xo = 13 m. Calculate the work done by this .
force acting on an object as it movesfrom x = 0                                          _ (2a) (x 3/2 _
- '3           2
3/2)
xl'
to x = 37 m. Hint: Consult Appendix A for the
integral of the cosine function and treat the                    Therefore, (a) WO-+3 = (2/3)(9.5 N/ml/2)(3 m)3/2                =
argument of the cosine as a quantity in radians.                 32.9 J, (b) W3-+6 = (6.33 N.m-1/2)[(6 m)3/2 -
(3 m)3/2] = 60.2 J, and (c) W6-+9 = 77.9 J.
Solution
From Equation 7-8:                                                   Problem
37m   32. A force given by F = b/..jXacts in the x direction,
W =
1
o
Fo cos -=-dx = Fo I xo sin
37m
Xo
-=- 1
xO o
where b is a constant with the units N.ml/2. Show
that even though the force becomes arbitrarily
= (51 N)(13 m) sin(37/13) = 193 J.                             large as x approaches 0, the work done in moving
from Xl to X2 remains finite even as Xl approaches
zero. Find an expression for that work in the limit
Xl -+ O.
Problem
30. Work the preceding problem graphically by                        Solution
making an accurate plot of the
force-versus-distance curve on graph paper, and
W =  1::  bX-I/2dx = 12bxl/21:: = 2b(y'x2 - y'xl),
which is finite for Xl -+ O. In fact, W = 2bJX2, for
determining the area (in units of work) under the
Xl -+ O.
curve.
Problem
33. The force exerted by a rubber band is given
approximately by

x                                             eo +x            e5]
F=Fo       [ e;;--        (iO+x)2         '
where eo is the unstretched length, X the stretch,
F;a>5 (X/x,,)                                   . and Fo is a constant (although Fo varies with
temperature). Find the work needed to stretch the
Problem 30 Solution.                                   rubber band a distance x.
CHAPTER7      113

parallel to dr along the path, then F.dr=Fldrl, and
W =F    I!   Idrl, where the scalar integral is now just
w =      t Fo [£0 £0
10
+     X' _       £~
(£0 + x')2
] dx'         the length of the semicircular path. Thus W = 1rRF.
(Note that the symbol dr is a displacement along the
Z

1 (       ,     X'2)    £~ I                  path, so Idrl is an element of path length, not the
= Fo 1 £0 lox       + 2" + £0 + x' 0                differential of the radius. Where confusion might arise,
2                                       one can use dl for path element, as in Chapters 25, 30,
=   FO(X+ 2£0 + ~
x
£o+x
-(0)'                and 31.)

(x' is a dummy variable)
(I)       +~
Section 7-4: Force and Work in                                                          A)                    (I)

Three Dimensions                                                                       ~      ---~--
(b) .••.•.•.•
~
Problem                                                                 It ,/                     I         "                 (b)

34. A car drives 3.1 kIn southward, propelled by the
I                    I                   .
6.3-kN force of friction between its tires and the                       { --                 I.                  I (c)

road. It then turns eastward and goes another                            •                               ~
l_ - - ...•. X
A.                                       B
1.8 kIn; during this stretch the frictional force is :'
5.1 kN. Finally, the car turns toward the southeast'                     FIGURE          7-29 Problem 35 Solution.
and goes 2.6 kIn while the frictional force is
6.8 kN. Find the total work done on the car by the
frictional force, assuming the force always acts in '       Problem
the direction of the car's motion.                          36. A cylindrical log of radius R lies half buried in the
ground, as shown in Fig. 7-30. An ant of mass m
Solution                                                            climbs to the top of the log. Show that the work
Since the frictional force is in the direction of motion            done by gravity on the ant is -mgR.
on each segment, this problem reduces to three
one-dimensional calculations: W = (6.3 kN)(3.1 km) +
(5.1 kN)(1.8km) + (6.8 kN)(2.6 km) = 46.4 MJ.

Problem
35. A particle moves from point A to point B along
the semicircular path of radius R, as shown in
Fig. 7-29. It is subject to a force of constant                               FIGURE         7-30 Problem 36.
magnitude F. Find the work done by the force
(a) if the force always points upward in Fig. 7-29,
(b) if the force always points to the right in
Fig. 7-29, and (c),if the force always points in the .
direction of the particle's motion.                         Solution
If we let dr=dxi+dyj,      as in Example 7-9 (wherei is
Solution                                                        horizontal to the right and j is vertical upward) then
(a) and (b) If the force is a constant vector (in              Wg = f~Fg .dr= f~(-mgj) • (dxi+dyj)         =
magnitude and direction) it may be factored out from           -my  f:~  dy = -mg(Y2 - Yl) = -mgR. (The difference
under the integral in the work (which is the limit of a   !i   in height going from the ground to the top of a half-
sum) to yield W =  I!   F • dr=F .       I!
dr. The          I'   buried log is just the radius.) Another way of
remaining integral is the sum of the displacements        'i   obtaining this result is to use reasoning similar to that
around a semicircle, which is just the total vector            in the solution to Problem 35. The force of gravity is
displacement along the diameter, from A to Bin                 constant and can be taken outside the integral, Wg =
Fig. 7-29. If we introduce 1ry coordinates to the right        I~ 9 • dr = F9 • Ii dr.The integral left is the total
F
and upward, respectively, with origin at the center of         displacement,     Ii
dr = Rj - Hi (if we take origin at the
the semicircle, then If dr=AB= 2Ri. For F=F1,                  center of the log's croSs-section), so Wg =
. W = Fj.2Ri="O, and for F=Fi,         W = Fi.2Ri=               (-mgj) .(Rj - Ri) = -mgR. Finally, one could use
2RF. (c) If the force is constant in magnitude and
114    CHAPTER 7

Equation 7-3 for the dot product, as shown:                                    the work becomes:
r2
Wg =   1    2
F 9 • dr = - 11            mg cos 0 Idrl                        W = .
1
(3'0)

(0,0)
F .i dx    +
/(3.6)

(3,0)
F .j dy

=-
r/
10
2
mgcosO.RdO=-mgRlsinOI~/
2
=1 = +1     3

Fz(Y          0) dx
6

Fy(x     = 3) dy

= +1 =
=-mgR.                                                                                              6
0            d dy       (15 N)(6 m)            = 90 J.
(Note: the x component of F on the first part of the
path, Fx(Y = 0), is xy = 0, and the y component of F
is tL)

d8                                                               Problem
39. You put your little sister (mass m) on a swing
whose chains have length i, and pull slowly back
until the swing makes an angle rjJwith the vertical.
Show that the work you do is mgi(l-     cosrjJ).

Problem 36 Solution.
Solution
Problem                                                                        The path is a circular arc (of radius i and differential
arc length Idrl = idO) so that only the tangential (or
37. A particle of mass m moves from the origin to the
parallel) components of any forces acting do work on
point x = 3 m, y = 6 m along the curve y =
the swing; the radial (or perpendicular) components do
ax2 - bx, where a = 2 m-I and b =4. It is subject
no work since the dot product with the path element
to a force F = cxyi + dj where c = 10 Njm2, and
is Zero. Thus, the tension in the chains and the radial
d = 15 N. Calculate the work done by the force.
components of gravity or the applied force do no work.
Solution                                                                        If you pull slowly, so that the tangential acceleration is
zero, then I'll = mgsinO, and the work you do is
Since the equation for the curve gives y as a function
of x, we can eliminate y and dy in the line integral for                                   rei>                    rei>,                 rei>
the work:                                                                          w= 10          F.dr=          10    FJildrl    = 10          mgsinO.£d(}

W =    lr~ r)
F . dr =         {(3.6)
1(0,0)
(Fxdx   + Fy dy)                   = mgi       1- cos Ol~ = mgi(l              - cos rjJ).
(Since forces perpendicular to the path of the swing do
=   1         3
[cx(ax   2
-    bx)   + d(2ax     - b)) dx            no work, all the work you do, for zero tangential

3                           2                 3
x4   'X
= c ( aT - b"3 )                     + d (x2aT     - bx )   10
I
= (45 + 90) J = 135 J,
where the given value!:!for the constants were used,
and all distances are in meters.

Problem
38. Repeat the preceding problem for the case when
the particle moves first along the x-axis from the
origin to the point (3,0), then parallel to the y-axis
until it reaches (3,6).

Solution
The path element dr equals idx for the first part of
the path and j dy for the second. The line integral for                                                  Problem 39 Solution.
CHAPTER 7        115

acceleration, is done against gravity, W = - Wg• But         Solution
w: =   -my Ay (as in Example 7-9), where Ay is the           We want !mcvb = !mTvf, so Vc         = vT/mT/mc =
ch~ge in height of the swing. In this case, Ay           =   (20 km/h) /(3.2xl04 kg)/(950 kg)      = 5.80VT =
£ _ £cosljJ, which gives the same Was above.)                116 km/h.

Section 7-5:            Kinetic Energy                       Problem
Problem                                                      44. A 60-kg skateboarder comes over the top of a hill
40. A 2.4x 105 kg jet is cruising at 900 km/h. What is           at 5.0 mis, and reaches 10 m/s at the bottom of
the kinetic energy relative to the ground? A 65-kg           the hill. Find the total work done on the
passenger strolls down the aisle at 3.1 km/h. What           skateboarder between the top and bottom of the
is the passenger's kinetic energy in the reference           hill.
frame of (a) the plane and (b) the ground?
Solution
Solution                                                     The work-energy theorem, Equation 7-16, gives
=
(a) Kjet = !(2.4x105 kg)(900 m/3.6 S)2 7.50 GJ (see,         Wnet = AK = ~m(v~ - v?) = !(60 kg) x
Table 1-1). (b) Relative to the plane,K~ass = !(65 kg),      (102 - 52)(m/s)   =2.25 kJ.          .
(3.1 m/3.6 s)2 = 24.1 J, but (c), relative to the ground
(the reference frame used in part (a)), the speed of the     Problem
passenger is 900:1: 3.1 lan/h, depending on his or her
45. Two unknown elementary particles pass through a
direction down the aisle. Therefore K pass           =           detection chamber. If they have the same kinetic
!(65 kg)(900:l: 3.1)2(1 m/3.6 s)2 .;" 2.05 MJ or
energy and their mass ratio is 4 : 1, what is the
2.02 MJ.
ratio of their speeds?

Problem                                                      Solution
41. Electrons in a color TV tube are accelerated to
25% of the speed of light. How much work does
Ifml == 4m2 and Kl == K2, then Vl == /2Kl/ml =
/m2vUml == v2/m2/ml == !V2; a mass ratio of 4 : 1
the TV tube do on each electron? (At this speed,
corresponds to a speed ratio of 1 : 2 if the kinetic
relativity introduces small but measurable
energy is the same.
corrections; here you neglect these effects.) See
inside front cover for the electron mass.
Problem
Solution                                                     46. You do 8.5 J of work to stretch a spring of spring
From ,the work-energy theorem, Wnet == AK ==                     constant k == 190 N/m, starting with the spring \
=
!me(0.25 C)2 0.5(9.11xlO-3l kg)(0.25x3x                          unstretched. How far does the spring stretch?
108 m/s)2 = 2.56xlO-15 J = 16.0 keV.
Solution
Problem                                                      The work done on a spring in stretching it a distance     x
42. In a cyclotron used to produce radioactive               from its unstretched length is (Equation 7-10 W =
isotopes for mediCal research, deuterium nuclei          !kx2. Thus x = /2W/k == 2(8.5 J)/(190 N/m) ==
(mass 3.3x 10-27 kg) are given kinetic energies of       29.9 cm. (See also Example 7-7(a).)
8.8xlO-l3J.   What is their speed? Compare with
the speed of light.                                      Problem
47. ~fter a tornado, a O.5G-g drinking straw was found
Solution                                                         embedded 4.5 em in a tree. Subsequent '

v   ==   J2K
m
=   [2(8.8XlO-
l3

3.3x 1O-27kg
J)]1/2
measurements showed that the tree would exert a
stopping force of 70 N on the straw. What was the
straw's speed when it hit the tree?
= 2.31x107 m/s        = 0.077c
Solution
Problem                                                      Since the stopping force (70 N) is so much larger than
43. At what speed must a 950-kg subcompact car be            the weight of the straw (0.0049 N), we may assume
moving-to have the same kinetic energy as a              that the net work done is essentially that done by just
3.2xI04_kg truck going 20 km/h?                          the stopping force, and use the work-energy theorem,
116    CHAPTER 7

Wnet = I:1K.The force is opposite to the displacement,      I:1K = O.The work done by gravity is Wg = mgx
so -F I:1r = 0 - !mv2, or v = v2Fl:1r/m =                   (72 cm), and that done by the stopping force is
V2(70 N)(0.045 m)/0.5xlO-3 kg = 112 m/s                     WF = -F(2 cm). Therefore, mg(72 cm) -
(- 250 mifh).                                               F(2 cm) = 0, or F = (8 kg)(9.8 mji)(72/2) =
2.82 kN = 635 lbs.
Problem
48. You drop a 15D-gbaseball from a sixth-story             Problem
window 16 m above the ground. What are (a) its          50. From what height would you have to drop a car
kinetic energy and (b) its speed when it hits the           for its impact to be equivalent to a collision at
ground? Neglect air resistance.                             20 mph?

Solution                                                    Solution
(b) The speed of the baseball (magnitude of the               "Equivalent" means the same kinetic energy is lost by
velocity) followsfrom Equation 2-11, the initial             the car as in the collision. Therefore, the impact speed
conditions (with y = 0 at ground level) and the neglect      is 20 mi/h = J29Ti, or h = (20x1.609 m/3.6 s)2+
of air resistance: v = V-2g(y - Yo) =                        2(9.8 m/s2) = 4.08 m = 13.4 ft;
J-2(9.8 m/s2)(-16 m) = 17.7 m/s. (a) The kinetic             Problem
energy is K = ~mv2 = !(0.150 kg)(17.7 m/s)2 =
51. Catapults run by high-pressure steam from the
23.5 J. Alternatively, K = ~m(2gyo) = mgyo, and
ship's nuclear reactor are used on the aircraft
v= v2K/m.                                                        carrier Enterprise to launch jet aircraft to takeoff
speed in only 76 m of deck space. A catapult
Problem                                                          exerts a 1.1x106 N force on a 3.3x104 kg aircraft.
49. A hospital patient's leg slipped off the stretcher           What are (a) the kinetic energy and (b) the speed
and his heel hit the concrete floor. As a physicist,        of the aircraft as it leaves the catapult? (c) How
estimate that the foot and leg had an effective             (d) What is the acceleration of the aircraft?
mass of 8 kg, that they dropped freely a distance
of 70 cm, and that the stopping distance was            Solution
2 cm. What force can you claim the floor exerted        (a) If we assume that the carrier deck and catapult
on the foot? Give your answer in pounds for the         force are horizontal, then the catapult force is the net
jury's sake.                                            force acting on the aircraft. The work-energy theorem
(Equation 7-16) gives I:1K = K - 0 = Wnet =
Fnetl:1x = (l.lx106 N)(76 m) = 83.6 MJ, for an
Solution                                                    aircraft starting from rest. b From Equation 7-15,
This problem can be solved with the use of Newton's         v = V2K/m = 2(83.6 MJ)/(3.3x104 kg) =
second law and kinematics (see Problem 5-67, for            71.2 m/s = 256 kID/h. (c) From Equation 2-9, t =
example), but alternatively, we may apply the               2(x - xo)/(vo + v) = 2(76 m)/(O + 71.2 m/s) = 2.14 s.
work-energy theorem: Wnet = Wg + W F = I:1K. The            (d) The acceleration can be calculated from Fnetlm =
leg starts from rest at point A and stops at point B, so    33.3 m/s2, or from Equation 2-7, a = vito In fact,
v/t = v/(2 I:1x/v) = v2/2 I:1x = (2K/m)/(2I:1x) =
Fnet/m.
\
Section 7-6:    Power
Problem
52. A horse plows a 20D-m-Iongfurrow in 5.0 min,
exerting a force of 750 N. What is its power
output; measured in watts and in horsepower?
Solution
If the force is assumed parallel to the displacement,
Problem 49 Solution.                       the average power is Pay = I:1W/l:1t= (750 N)x
(200 m)j(300 s) = 500 W = (500/745.7)hp = 0.671 hp.
CHAPTER7       117

Problem                                                            launch? (b) If the driver makes one launch every
53. A typical car battery stores about 1 kWh of                    30 min, what is its average power consumption?
energy. What is its power output if it is drained
completely in (a) 1 minutej (b) 1 hourj (c) 1 day?         Solution
(a) The work done by the driver during a launch is
Solution                                                       equal to the change in kinetic energy of the package,
The average power (Equation 7-17) is (a) P =                   so the power output is P = Wit = D.Klt =
(1 kWh)/(1 h/60) = 60 kWj (b) 1 kWh/1 h =                      ~(103 kg)(2 km/s)2/55 s= 36.4 MW. (b) The same
1 kWj and (c) 1 kWh/24 h = 41.7 W,                             work (one launch) spread over 30 min yields an
average power of Pay = D. WI D.t = 2 x 109 J ..;-
Problem                                                        30x60 s = 1.11 MW. (Note: we are neglecting any
work done by lunar gravity during the launch, which
54. A sprinter completes a 1OD-mdash in 10.6 s, doing
would be a small c~rrection to W.)
22.4kJ of work. What is her average power
output?
Problem
Solution                                                       58. A-75-kglong-jumper  takes 3.1 s to reach a
prejump speed of 10 m/s. What is his power
From Equation 7-17, P = D.W/D.t = 22.4 kJ/10.6 s =
output?
2.11 kW. (The distance of the dash is not relevant
here.)
Solution
Problem                                                        P = Wit  = D.K/t = !(75 kg)(l0 m/s)2/3.1     s=
1.21 kW ~ 1.6 hp.
55. How much work can a 3.5-hp lawnmower engine
do in 1 h?
Problem
Solution                                                       59. Estimate your power output as you do deep knee
Working at constant power output, Equation 7-17                    bends at the rate of one per second.
gives the total work (energy output) as D.W =
P D.t = (3.5 hp) (746 W/hp)(3600 s) = 9.40 MJ.                 Solution
(Note the change to appropriate SI units.)                     The work done against gravity in raising or lowering a
weight through a height, h, has magnitude mgh. The
Problem                                                        body begins and ends each deep knee bend at rest
56. Water drops over 49-m-high Niagara Falls at the            (D.K = 0), so the muscles do a total work (down and
rate of 6.0 x 106 kg/so If all the energy of the falling   up) of 2mgh for each complete repetition. If we
water could be harnessed by a hydroelectric power,:        assume that the lower extremities comprise 35% of the
plant, what would be the plant's power output?             body mass, and are not included in the moving mass,
then mg, for a 75 kg person, is about 0.65(75 kg) x
2
Solution                                                       (9.8 m/s ) = 480 N. We guess that h is somewhat
greater than 25% of the body height, or about 45 em,
The force of gravity does work dWg = -(dm)g D.y on
so the muscle power output for one repetition per
each mass dm of water going over the falls. If we
second is about 2(480 N)(0.45 m)/s =430 W.
neglect any energy associated with the current flow,
this. is the energy that could be harnessed by a
hydroelectric plant. Since the'rate of flow, dm/dt,
Problem
\
is given, the available power can be found: P =                60. At what rate can a one-half horsepower well pump
dWg/dt = -(dm/dt)g D.y = -(6x106 kg/s)x                            deliver water to a tank 60 m above the water level
2
(9.8 m/s )( -49 m) = 2.88 GW.                                      in the well? Give your answer in kg/s and gal/min.

Problem                                                        Solution
57. A "mass driver" is designed to launch raw                  If D.m/ D.t is the mass of water pumped per second,
material mined on the moon to a factory in lunar           the work done (lifting D.m against gravity at constant
orbit. The driver can accelerate a 100D-kg package         speed) per second is P = D.W/D.t = (D.m/D.t)gh.
to 2.0-km/s Uust under lunar escape speed) in              Therefore, D.m/D.t = P/gh = (0.5 hp)(746 W/hp)+
55 S. (a) What is its power output during a                (9.8 m/s2)(BO m) = 0.634 kg/so From Appendix C,
118   CHAPTER 7

3
the density of water is (103 kg/m )(3.786x                 Problem
1O-3m3/gal) = 3.786 kg/gal, so 6.m/6.t    =                65. By measuring oxygen uptake, sports physiologists
(0.634 kg/s)(60 s/ min)/(3.786 kg/gal) = 10.1 gal/min.         have found that the power output of long-distance
runners is given approximately by P = m(lw - e),
Problem                                                        where m and v are the runner's mass and speed,
61. In midday sunshine, solar energy strikes Earth at          respectively, and where band c are constants
2
the rate of about 1 kW /m • How long would it              given by b = 4.27 J/kg.m and e = 1.83 W /kg.
take a perfectly efficient solar collector of 15.m 2       Determine the work done by a 54-kg runner who
area to collect 40 kWh of energy? (This is roughly         runs a 100km race at a speed of 5.2 m/s.
the en.ergy content of a gallon of gasoline.)
Solution
Solution                                                   The runner's average power output is P = (54 kg) x
The average power received by the collector is             [(4.27 J/kg.m)(5.2 m/s) -1.83 W/kg) = 1.10 kW.
(1 kW /m2)(15 m2) = 15kW, so it would take                 Over the race time, 10 km/(5.2 m/s) = 1.92x 103 s,
6.t = 6.W/P = 40 kWh/15 kW = 2.67 h to collect the         the runner's work output is 6.W = P 6.t = 2.12 MJ      =
required energy. (The average intensity of sunlight is     0.588 kWh.
discussed in more detail in Chapter 34.)
Problem
Problem                                                    66. A 65-kg runner running at Vo = 4.8 m/s
62. It takes about 20 kJ to melt an ice cube. A typical        accelerates to 6.1 m/s over a 25-s interval. (a) By
microwave oven produces 625 W of microwave                 writing v = Vo + at, where a is the runner's
power. How long will it take to melt the ice cube          acceleration, use the formula in the previous
in this oven?                                              problem to express the runner's power output as a
function of time. (b) How much work does the
Solution                                                       runner do during the acceleration period?

From Equation 7-17,6.t   = 6.W/P = 20 kJ/625     W    =    Solution
32.0 s.
(a) P(t) = m(bvo + bat - e). (b) W = f~' P(t) dt =
f~,=25S m(lwo + bat - e) dt = m(lwotl + !bat~ - ett} =
Problem
mtl[!b(vo + Vl)- c) = (65 kg)(25 s)[!(4.27 J/kg.m)
63. The rate at which the United States imports oil,
(4.8   + 6.1)(m/s)   - 1.83 W Ikg) = 34.8 kJ.
expressed in terms of the energy content of the
imported oil, is nearly 700 GW. Using the
Problem
"Energy Content of Fuels" table in Appendix C,
67. A 1400-kg car ascends a mountain road at a
convert this figure to gallons per day.
steady 60 km/h. The force of air resistance on the
car is 450 N. If the car's engine supplies energy to
Solution
the drive wheels at the rate of 38 kW, what is the
Appendix C lists the energy content of oil as
39 kWh/gal. Therefore; the import rate is
(700 GW)(1 gal/39 kWh) x (24 hid), or roughly              Solution
430 million gallons per day.
At constant velocity, there is no change in kinetic
energy, so the net work done on the car is zero.
Problem
Therefore, the power supplied by the engine equals the
64. Which consumes more energy, a 1.2-kW hair dryer        power expended against gravity and air resistance.
used for 10 min or a 7-W night light left on for       The latter can be found from Equation 7-21, since
24 h?                                                  gravity makes an. angle of () + 90° with the velocity
(where () is the slope angle to the horizontal), while air
Solution                                                   resistance makes an angle of 180° to the velocity. Then
The energy consumption of each device is 6.W =             38 kW = -Fg -V-Fair -v = -mgvcos«(} + 90°) -
P6.t; (1.2 kW)(1 h/6) = 0.2 kWh for the hair dryer         Fair V cos(1800) = mgv sin () + Fair v, or () = sin -1 x
and slightly less, (7 W)(24 h) = 0.168 kWh, for the        [«38 kW 160 km/h) - 450 N)/(1400 kgx
night light:                                               9.8 m/s2») = 7.67°. (See Example 7-14 and use care
with SI units and prefixes.)
CHAPTER 7       119

Problem                                                                 Problem
2
68. A machine does work at a rate given by P =             ct ,     •   72. A force pointing in the x direction is given by F =
where c = 18 W /S2, and t is time. Find ~n                  .           ax3/2, where a == 0.75 N/m3/2. Find the work
expression for the work done by the machine                             done by this force as it acts on an object moving
between t = 10 s and t = 20 s.                                          from x = 0 to x = 14 m.

Solution                                                                Solution
Equation 7-18 implies                                                                       14m               5/2/l4m

W=
1
205 1205 dt
Pdt
105
ct =
lOs
2
W =
1o
ax3/2 dx = aX
5
(2)     0

= ~ (0 75~)      (14 m)5/2 = 220 J
5    . m3/2
= ~ (18 ~)   [(20 s)3 - (10 s)3]

=42 kJ.
Problem
Paired Problems                                                         73. Two vectors have equal magnitude, and their
scalar product is one-third of the square of their
Problem
magnitude. Find the angle between them.
69. You apply a 470-N force to push a stalled car at a.
17° angle to its direction of motion, doing 860 J of                Solution
work in the process. How far do you push the car?
We are given that A2 = B2 = 3A.B          = 3ABcos().
Therefore () = cos-l(I/3) = 70.5°.
Solution
Equation 7-2 or 7-5 gives the work done by the applied
Problem
force; hence !:i.r = W/Fcos() = 860 J/(470 Nx
74. Vector A has magnitude A, vector B has
cos 17°) = 1.91 m.
magnitude 2A, and A. B = A 2• Find the angle
Problem                                                                     between A and B.

70. A tractor tows a jumbo jet from its airport gate,
Solution
doing 8.7 MJ of work. The link from the plane to
the tractor makes a 22° angle with the direction of
A2 = A. B = AB cos () = 2A2 cos ()j therefore () =
the plane's motion, and the tension in the link is                  cos-l(1/2)   = 60°.
4.1xl05 N. How far does the tractor move the
plane?                                                              Problem
75. A 460-kg piano is pushed at constant speed up a
Solution                                                                    ramp, raising it a vertical distance of 1.9 m (see
As in the previous problem,                                                 Fig. 7-32). If the coefficient of friction between
!:i.r = 8.7 MJ/(4.lxl05 Nx cos 22°) = 22.9 m.

Problem
71. A force pointing in the x direction is given by
F = Fo(x/xo), where Fo and Xo are constants, and
x is the position. Find an expression for the work
done by this force as it acts on an object moving
from x = 0 to x = Xo.

Solution
The work done in this one-dimensional          situation   is
given by Equation 7-8:

W = rO(Fo)x          dx = (Fo)    x~ = ~Foxo.
10      Xo             Xo       2      2                                     FIGURE 7-32 Problem       75.
120     CHAPTER 7

ramp and piano is 0.62, find the work done by the     Problem
agent pushing the piano if the ramp angle is          77. (a) How much power is needed to push a 95-kg
(a) 15° and (b) 30°. Assume the force is applied ._       chest at 0.62 m/s along a horizontal floor where
parallel to the ramp.                                                                     7
the coefficient of friction is O. 8? (b) How much
work is done in pushing the chest 11 m?
Solution
The usual relevant forces on an object pushed up an "      Solution
incline of length l = hi sin 9 by an applied force         If you push parallel to the floor at constant velocity,
parallel to the slope are shown in the sketch. At          the normal force on the chest equals its weight,
constant velocity, the acceleration is zero, so the        N = mg, and the applied force equals the frictional
parallel and perpendicular components of Newton's          force, Fapp =!k = ILkN = ILkmg. (a) The power
second law, together with the empirical relation for       required is (Equation 7-21) Papp = Fappv = (0.78)x
kinetic friction, give N = mgcos9, fk = ILkN, and          (95 kg)(9.8m/s2)(0.62 m/s) = 450 W, or about
Fapp = mg sin9 + fk = mg(sin 9 + ILkcos 9). Thus, the      0.6 hp. (b) The work done by the applied force acting
work done by the applied force is Wapp = Fappl =           over a displacement tlx = 11 In is Wapp = Fapp.6.x =
Fapphl sin 9 = mgh(1 + ILkcot 9), where h is the           Papp(.6.xlv), where t = .6.xlv is the time over which
- vertical rise. (a) Evaluating the above expression using   the power is applied. Using either expression, we find
the data supplied, we find Wapp = (460 kg) x               Wapp = 7.99 kJ.
(9.8 m/s2)(1.9 m)(1 + 0.62 cot 15°) = 28.4 kJ.
(b) When 15° is replaced by 30° in the above               Problem
calculatiion,we find Wapp = 17.8 kJ. (The work done        78. You mix flour into a thick bread dough, exerting a
against gravity is the same in parts (a) and (b) since h       45-N force on the stirring spoon. If you move the
is the same, but the work done against friction is             spoon at 0.29 mis, (a) what power do you supply?
greater in (a) because the incline is longer and the           (b) How much work do you do if you stir for
normal force is greater; -however Fapp is less.)               1.0 min?

-
i
Solution
(a) Provided the stirring force is applied always
parallel to the velocity of the spoon, Papp =
h             Fappv = (45 N)(0.29 m/s) = 13.1 W. (b) W~pp =
Pappt = (13.1 W)(60 s) = 783 J.

J             Supplementary Problems
Problem 75 Solution.                 Problem
79. The power output of a machine of mass m
increases linearly with time, according to the
Problem                                                         formula P = bt, where b is a constant. (a) Find an
76. You have to do 2.2 kJ of work to push a 78-kg               expression for the work done between t = 0 and
trunk 3.1 m along a slope inclined upward at 22°,           some arbitrary time t. (b) Suppose the machine is
pushing parallel to the slope. What is the                  initially at rest and all the work it supplies goes
coefficientof friction between trunk and slope.             into increasing its own speed. Use the work-energy
theorem to show that the speed increases linearly
Solution                                                         'with time, and find an expression for the
Using the equation derived for Wapp in the previous              acceleration.
problem, one can solve for the coefficient of friction:
Solution
ILk =   (W app
mgh
-   1) tan 9                         (a) From Equation 7-'20,W = J~Pdt' = J~bt' dt' =
!bt2. (We used t' for the dummy variable of
integration.) (b) If we assume that W = Wnet = .6.K,
_ (            2.2kJ               -1)             then !bt2 = ~mv2, since the machine starts from rest.
- (78 kg)(9.8 m/s2)(3.1 m) sin 22°
Thus v = Jblm t and a = dvldt = Jblm. (v is the
x tan 22° = 0.60.                             speed -and a is the tangential acceleration along the
path of the machine.)
CHAPTER 7          121

em                                                       Solution
ou're trying to decide whether to buy an                If we ignore the mass of the spring, the spring tension
ergy-efficient,225-W refrigerator for \$1150 or a       increases from 0 to mg in the interval before the mass
dard, 425-W model for \$850. The standard             leaves the ground. The work done against the spring
..odel will run 20% of the time, while better            force is W =    I::
Fsprdx, as in Example 7-7. Since we               :"]
'Jl
ulation means the energy-efficient model will         know Fspr as a function of X, Fspr = kx, we can find
irun 11% of the time. If electricity costs 9.5t/kWh,        the appropriate limits for the interval Xl = 0 to X2 =
how long would you have to own the energy-                mg Ik and integrate over x. The result is W =
','efficient odel to make up the difference in cost?
m                                               {:9/k kx dx = !k(mglk)2 = m2g2/2k. (Alternatively,
'Neglect interest you might earn on your money.             we can use Fspr = kx to eliminate X, and integrate
over Fspr as a variable. This gives dFspr = k dx and
uti on                                                     W = 1;"9 Fspr(dFsprlk) = Hmg)2Ik.)
e price differential is equal to the difference in
cost of energy over a period of time t, then               Problem
(1150- 850) = (Pstdx20%t - PeffX l1%t) x                     83. Figure 7-33 shows the power a baseball bat
\$0.095/kWh). Solving for t, we find t = (300)x                    delivers to the ball, as a function of time. Use
[(0.425xO.2- 0.225xO.11)(0.095)]-1 h =                             graphical integration to determine the total work
5.24x104 h = 2.18x103 d ~ 6.0 y.                                   the bat does on the ball.

.Problem
81. The per-capita energy consumption rate plotted in
Fig. 7-21 can be approximated by the expression
P = Po +at + bt2 + et3, where Po = 4.4 kW,
a = -5.57xl0-2 kW Iy, b = 3.84xlO-3 kW ly2,
C = -2.79xlO-5     kW Il, and t is the time in years
since 1900 (i.e., 1960 is t = 60). Integrate this
expression to find approximate values for the                               FIGURE       7-33 Problem 83.
energy used per capita during the decades
(a) from 1940 to 1950 and (b) from 1960 to 1970.
The work done on the ball, W = 1 Pdt, is the area
Solution                                                       under the graph of power versus time, where each
The energy used between times h and          t2   is           small rectangle in Figure 7-33 has an "area" of

W =    {t p dt
1tl
2

=   I
tl
h
(Po + at   + bt2 + ct3)dt
(0.01 s)(1 kw) = 10 J. There are approximately 54 or
55 rectangles under the curve, so the work done was
= PO(t2 - h)   + !a(t~     - tn   + ib(t~      - t~)
+ic(t~ -t1),                                        Problem
84. A machine delivers power at a decreasing rate
(a) Using tl = 40 y, t2 = 50 y, and the given
P = PotV(t + to)2, where Po and to are constants.
coefficients,we find the energy used in the 1940s to be
The machine starts at t = 0 and runs forever.
W = 71.3 kW.y. (b) A similar calculation for the
Show that it nevertheless does only a finite
1960sgives W = 93.3 kW.y.
amount of work, equal to Poto.
Problem                                                        Solution
82. A spring of spring constant k is attached to a mass        From Equation 7-20,
m, and the other end of the spring is pulled
vertically in order to lift the mass. Find an
expression for the amount of work that must be
W~    r~)Pot5 dt
10     (t + to)2
=_   I(t+ to) /00
Pot5
0
= Poto.

done on the spring before the mass begins to leave
the ground.                                                Problem
85. An unusual spring has the force-distance curve
shown in Fig. 7-34 and described by F = 100x2 for

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