VIEWS: 0 PAGES: 15 CATEGORY: Technology POSTED ON: 3/9/2010
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 you are called to testify about this accident. You long does the catapulting operation take? 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 slope angle of the road? 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. It's easiest to give your answer in kilowatt-years. Solution 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 about 545 J. = 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