# Microsoft PowerPoint - Ch 7 r.pp

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```					                      Chapter 7
Units of Chapter 7
Work and Energy
Work Done by a Constant Force
Scalar Product of Two Vectors
Work Done by a Varying Force
Kinetic Energy and the Work-Energy Principle

1                                                                 2

7-1 Work Done by a Constant Force                                            7-1 Work Done by a Constant Force
The work done by a constant force is defined as the distance moved
multiplied by the component of the force in the direction of displacement:

As long as this person does not
lift or lower the bag of groceries,
In the SI system, the units of work are joules:                                                    he is doing no work on it.
The force he exerts has no
component in the direction of
motion.

3                                                                 4
7-1 Work Done by a Constant Force                                                           7-1 Work Done by a Constant Force

Example 7-2, page 166.
(a) Determine the work a hiker must do on
Solving work problems:                                                                a 15.0-kg backpack to carry it up a hill of
1. Draw a free-body diagram.                                                                            height h = 10.0 m, as shown. Determine
also (b) the work done by gravity on the
2. Choose a coordinate system.                                                                          backpack, and (c) the net work done on
3. Apply Newton’s laws to determine any unknown                                                         the backpack. For simplicity, assume the
forces.                                                                                              motion is smooth and at constant velocity
(i.e., acceleration is zero).
4. Find the work done by a specific force.
5. To find the net work, either
SOLUTION:
a) find the net force and then find the work it does, or                                             1. Draw a FBD and show forces acting on
the backpack.
b) find the work done by each force and add.
2. Choose a coordinate system.
5                                                                       6

7-1 Work Done by a Constant Force                                      7-2 Scalar Product of Two Vectors
3. Apply Newton’s laws.              ∑Fy = may      →     FH – mg = 0                Definition of the scalar, or dot, product:
Hence FH = mg = (15.0 kg)(9.8 m/s2) = 147 N.
4. Work done by a specific force. (a) Work done by the hiker on the
backpack is      WH = FH (d cosθ) = FH h = mgh = (147 N)(10.0 m) = 1470 J.

(b) The work done by the gravity on the
backpack is WG = FG d cos(180° - θ)
WG = FG d(- cosθ) = mg(-d cosθ) = - mgh
= - (15.0 kg)(9.8 m/s2)(10.0 m) = - 1470 J.

Therefore, we can write:
5. Net work done. (c) The net work done on the pack is
Wnet = WG + WH = - 1470 J + 1470 J = 0.
Note: Even though the net work done by all the forces on the backpack is zero,
the hiker does do work on the backpack equal to 1470 J.                          7                                                                       8
7-2 Scalar Product of Two Vectors                                             7-2 Scalar Product of Two Vectors
A·B=B·A                           Commutative              Example 7-4: The force shown has magnitude FP = 20 N and makes
an angle of 30° to the ground. Calculate the work done by this force,
A · (B + C) = A · B + A · C                Distributive
using the dot product, when the wagon is dragged 100 m along the
ground.
Unit vectors      i·i= j·j=k·k=1
i·j=i·k=j·k=0

If      A = Ax i + Ay j + Az k
B = Bx i + By j + Bz k
then     A · B = AxBx + AyBy + AzBz                                           Solution:       W = FP · d

FP = (FP cos 30°) i + (FP sin 30°) j = 17 N i + 10 N j and d = (100 m) i
Special case A = B
W = FP · d = (17 N)(100 m) + (10 N)(0) = 1700 J.
A · A = Ax 2 + Ay 2 + Az 2 = A2 .
9                                                                            10

7-3 Work Done by a Varying Force                                              7-3 Work Done by a Varying Force
Particle acted on by a varying force. Clearly, F · d is not constant!

For a force that varies, the work can be approximated by dividing the
distance up into small pieces, finding the work done during each, and
adding them up. The work done ΔWi by the force Fi for the i-th interval is

Δwi ≈ Fi cosθi Δℓi
11                                                                            12
7-3 Work Done by a Varying Force                                         7-3 Work Done by a Varying Force
In the limit that the pieces become infinitesimally narrow, the                                                Work done by a spring force:
work is the area under the curve:
The force exerted by a spring is
given by Hooke’s law:

or:                                                                                                                                  .
k is called the spring constant.

Note the negative sign…

Force exerted by the spring is
always directed opposite the
displacement.
13                                                                                14

7-3 Work Done by a Varying Force                                       7-4 Kinetic Energy and the Work-Energy Principle

If we write the acceleration in terms of the velocity and the distance,
Plot of F vs. x.                      we find that the work done is
Work done is equal to the                                           ⎛ v2 − v12 ⎞
2
⎛ v2 − v12 ⎞
2

shaded area.                               Wnet                     ⎜ 2d ⎟ d = m⎜ 2 ⎟
= Fnet d = mad = m⎜          ⎟ ⎜          ⎟
⎝          ⎠ ⎝          ⎠
or

We define the quantity mv2/2 to be the translational kinetic
energy K of the object:

15                                                                                16
7-4 Kinetic Energy and the Work-Energy Principle                                                       Problem 49 (3rd ed.)
A force of 6.0 N is used to accelerate a mass of 1.0 kg from rest for a
This means that the work done is equal to the change in the                      distance of 12 m. The force is applied along the direction of travel. The
kinetic energy:                                                                  coefficient of kinetic friction is 0.30. What is the work done (a) by the
applied force? (b) by friction? (c) What is the kinetic energy at the 12-m
mark?

Solution: (a) The work done by the applied force is
WF = F d = (6.0 N)(12 m) = 72 J
• If the net work is positive, the kinetic energy increases.
(b) The work by friction is
• If the net work is negative, the kinetic energy decreases.                                                  Wfr = FN d cos 180° = μk mg d (-1)
= - (0.30)(1.0 kg)(9.8 m/s2)(12 m) = -35 J
Because work and kinetic energy can be equated, they must                       (c) The normal force and weight do no work. The net work increases the kinetic
have the same units: kinetic energy is measured in joules.                      energy of the mass:
Wnet = WF + Wfr = ΔK
72 J – 35 J = Kf – 0 which gives       Kf = 37 J.
17                                                                                 18

Problems 47 (3rd ed.)                                                                       Problem 11
At an accident scene on a level road, investigators measured a car’s             A 380-kg piano slides
skid mark to be 78 m long. It was a rainy day and the coefficient of             3.9m down a 27°
kinetic friction was estimated to be 0.38. Use these data to determine           incline and is kept from
the speed of the car when the driver slammed on (and locked) the                 accelerating by a man
brakes?                                                                          who is pushing back
on it parallel to the
Solution:                                                                         incline. Determine: (a)
On a level road, the normal force is mg, so the kinetic friction force is         the force exerted by
μkmg. We have                                                                     the man, (b) the work
done by the man on
W = ΔK →      μkmg d cos 180° = ½ mv2 - ½ mv0 2                       the piano, (c) the work
done by the force of
(0.38) m (9.8) (78) (-1) = - ½ mv0 2                         gravity, and (d) the net
work done on the
v0 = 24 m/s            (87 km/h)         piano. Ignore friction.

19                                                                                 20
End of Problem 11 # 11
Chapter Problems                                                                       chapter problems
End ofProblem 11 # 11
r                                                           .

FN                                                    (b)
WP = FPd cos180° = − (1691N)( 3.9m) = −6595J ≈ −6600J
r
(a)                                 FP         y
θ

(      )
(c)
W = F d cos63° = mgd cos63° = ( 380kg) 9.80m s2 ( 3.9m) cos63°
r
mg                   x                                      G   G

= 6594N ≈ 6600J
∑F  y
= FN − mg cos θ = 0 → FN = mg cos θ

∑F  x
= mg sin θ − FP = 0 →                                            (d) Since the piano is not accelerating, the net force on the piano
is 0, and so the net work done on the piano is also 0. This
FP = mg sin θ = mg sin θ                                                    can also be seen by adding the two work amounts calculated.

(
= ( 380 kg ) 9.80 m s2    ) ( sin 27°) = 1691N ≈ 1700 N                               Wnet = WP + WG = −6.6 × 103 J + 6.6 × 103 J = 0 J
21                                                                               22

End ofProblem 13 # 13
chapter problems                                                                 End ofProblem 13 # 13
chapter problems

A 17,000-kg jet takes off from
Solution:
an aircraft carrier via a
catapult. The gasses thrust out       (a) The gases exert a force on the jet in the same direction as the
of jet’s engines exert an                 displacement of the jet. From the graph we see the displacement
constant force of 130 kN on the           of the jet during launch is 85 m.
jet; the force exerted on the jet
by the catapult is plotted in                                                   (          )
Wgas = Fgas d cos 0° = 130 × 103 N ( 85m ) = 1.1 × 107 J
figure (b).

Determine:
(a)The work done on the jet by        (b) The work done by catapult is the area underneath the graph in
the gases expelled by its                 Figure 7-22. That area is a trapezoid.
engines during the launch of
the jet; and
(b)The work done on the jet by
Wcatapult =   1
2   (1100 × 10 N + 65 × 10 N ) (85m) = 5.0 × 10 J
3          3                      7

the catapult during launch of
the jet.
23                                                                               24
End of Problem 54 # 54
chapter problems                                                          Chapter Problems
End of Problem 54 # 54

First find the kinetic energy of the train, and then find out how
Spiderman uses his                                                     much work the web must do to stop the train.
spider webs to save a
runaway train,
Note that the web does negative work, since the force is in the
moving at 20 m/s. His
opposite direction of the displacement.
web stretches 800 m
before the 104 kg
train comes to a stop.
Assuming the web
Wto stop = ΔK = 1 mv2 − 1 mv12 = 0 − 1 104 kg ( 20 m s ) = −2 × 106 J
train
2
2
2            2      (         )                2

acts like a spring,
estimate the spring
constant.
Wweb = − 1 kx 2 = −2 × 10 6 J → k =
(        )=
2 2 × 10 6 J
6N m
2
(800 m )     2

25                                                                                     26

End of Problem 63 # 63
Chapter Problems                                                   End of Problem 63 # 63
Chapter Problems
(a) The angle between the pushing force and the displacement is 32°.
(a) How much work is done by the horizontal force FP = 150 N on the
18-kg block when the force pushes the block 5.0 m up along the 32°               WP = FPd cos θ = (150 N )( 5.0 m) cos32° = 636.0J ≈ 640J
frictionless incline? (b) How much work is done by the gravitational
force on the block during this displacement? (c) How much work is      (b) The angle between the force of gravity and the displacement is 122°.
done by the normal force? (d) What is the speed of the block (assume
that it is zero initially) after this displacement?                                                       (            )
WG = FGd cosθ = mgd cosθ = (18kg) 9.80m s2 ( 5.0m) cos122° = −467.4J ≈ −470J

(c) Because the normal force is perpendicular to the displacement, the
work done by the normal force is 0.

(d) The net work done is the change in kinetic energy.

W = WP + Wg + WN = ΔK =           1
2
mv2 −
f
1
2
m v i2   →

2W        2 ( 636.0 J − 467.4 J )
vf =          =                                     = 4.3 m s
27
m                  (18 kg )                                   28

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