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Unformatted text preview: CHAPTER 7 WORK, ENERGY, AND POWER
How much work do you do on the barbell during this time? (c) You lower the barbell to the ground. Now how much work do you do on it? ActivPhysics can help with these problems: Activity 5.1 Section 71: Work Problem
1. How much work do you do as you exert a 75N force to push a shopping cart through a 12mlong supermarket aisle? Solution
(a) If we assume the barbell is lifted at a constant velocity by a vertical applied force equal to the weight and positive upward parallel to the displacement, then Wapp == mg ~y == (45 kg)(9.8 m/s2)(2.5 m) == 1.10 kJ. (b) 'Just holding the weight stationary, you must still exert an applied force of my to balance gravity, but the displacement through which the force acts is zero. Hence the work done on the barbell is also zero. (Actually, individual muscle fibers are continually contracting even though the overall muscles are stationary, so internal work is being done in your muscles and you feel tired just holding a weight.) (c) When the weight is lowered at constant velocity, the upward applied force, mg, is opposite to the displacement downward, ~y == 2.5 m; hence the work done on the barbell is negative, Wapp == 1.10 kJ. Solution
If the force is constant and parallel to the displacement, W=F.~r=F~r==(75N)(12m) == 900J. Problem
2. If the coefficientof kinetic friction is 0.21, how much work do you do when you slide a 50kg box at constant speed across a 4.8mwide room? Solution
If you push parallel to a level floor, the applied force equals the frictional force (since the acceleration is zero), and the normal force equals the weight. The applied force is constant and parallel to the displacement, so Wa == Fa ~r == !k ~r == ItkN ~r == 2 Itkmg ~r == 0.21(50 kg)(9.8 m/s )(4.8 m) == 494 J. Problem
5. The world's highest waterfall, theCherunMeru in Venezuela, has a total drop of 980 m. How much work does gravity do on a cubic meter of water dropping down the CherunMeru? Problem
3. A crane lifts a 650kg beam vertically upward 23 m, then swings it eastward 18 m. How much work does the crane do? Neglect friction, and assume the beam moves with constant speed? Solution
The force of gravity at the Earth's surface on a cubic meter of water is Fg == mg== 9.8 kN vertically downward (see the inside book cover of the text for the density of water). The displacement of the water is parallel to this, so the work done by gravity on the water is Wg == Fg ~y == (9.8 kN)(980 m) == 9.6 MJ. Solution
Lifting the beam at constant speed, the crane exerts a constant force vertically upward and. equal in magnitude to the weight of the beam. During the horizontal swing, the force is the same, but is perpendicular to the displacement. The work done is : F. ~r == (mgj) .(~yj + ~xi) == mg ~y == (650 kg) x , (9.8 m/s2)(23 m) == 147 kJ. Pr6blem
6. A meteorite plunges to Earth, embedding itself 75cm in the ground. If it does 140 MJ of work in the process, what average force does the meteorite exert on the ground? Problem
4. You lift a 45kg barbell from the ground to a height of 2.5 m. (a) How much work do you do on the barbell? (b) You hold the barbell aloft for 2.0 min. Solution
The average force exerted by the meteorite parallel to its penetration into the ground is Fav == W/ ~x == 140 MJ/O.75 m == 187 MN, or about 21,000 tons. 108 CHAPTER 7 Problem
7. You slide a box of books at constant speed up a 30° ramp, applying a force of 200 N directed up the. slope. The coefficient of sliding friction is 0.18. (a) How much work have you done when the box has risen 1 m vertically? (b) What is the mass of the box? Problem
9. A locomotive does 7.9 x 1011 J of work in pulling a 3.4x105kg train 180 km. What is the average force in the coupling between the locomotive and the rest of the train? 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
mass is not required to answer this question.) Solution
(a) The displacement up the ramp (parallel to the applied force) is !1r = 1 m/ sin 30° = 2 m, so Wa = Fa' ~r= (200 N)(2 m) = 400 J. (b) We could easily solve Newton's second law, with zero acceleration, to find the mass, m = Fa/ g(sin (J + fJ.k cos (J), but it is instructive to obtain the same result using the concept of work. The work done by gravity is F 9 • ~r Fg•(!1xi+ !1yj)= mg !1y= m(9.8 m/s2)(l m) m(9.8 J /kg) (see Example 79). The work done by friction is fk • ~r =  fk!1r = fJ.kN !1.r = J1.kx (mg cos 3()O)!1.r = 0.18m(9.8 m/s2)(2 m) cos 30° = m(3.06 .J/kg). The total work is zero (v is constant in the workenergy theorem), so 0= 400 J  mx (9.8 + 3.06) J/kg, of m 400 J/(12.86 Jjkg) = 31.1 kg. Problem
10. An elevator of mass m rises a distance h up a vertical shaft with upward acceleration equal to onetenth g. How much work does the elevator cable do on the elevator? = = Solution
To give the elevator a constant upward acceleration all = 0.1 g, the tension in the cable must satisfy T ~ mg mall' or T m(g + all) = 1.1mg. Acting over a parallel displacement !1y = h upward, the tension does work WT = T!1.y 1.1 mgh on the elevator. = = = = Section 72: 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
8. Two people push a stalled car at its front doors, each applying a 280N force at 25° to the forward direction, as shown in Fig. 725. How much work does each do in pushing the car 5.6 m? This follows easily from the definition of the scalar product in terms of components: A. (B + C) = Ax(Bx + Gx) + AII(BIi + Gil) + Az(Bz + Gz) = AxBx AIiBIi + + AzBz + AxGx + AIiGIi + AzGz = A . B + A. C.
 (JB)
(law of sines), (law of cosines), With more effort, it also follows from trigonometry. First: Dsin«(JD  (JB) = Gsin(ec D = JB2 + G2 + 2BGcos«(Jc  (JB) FIGURE 725 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: together give D cos(9v  9B) == B Second: A.D=ADcos9v =AD[cos(9v  9B) COS9B  sin(9v  9B) sin9B] A has magnitude 10 and points 30° above the :rraxis B has magnitude 4.0 and points 10° to the left of the yaxis C = 5.6i  3.1j D = 1.9i 7.2j, compute the scalar products (a) A. B (b) C  D (c)BC. . + G cos(9c  9B)' + =A[B + Gcos(9c 9B)]cos9B A[Gsin(9c  9B)] sin9B =ABcos9B +AG{cos(9c  9B)cos9B  sin(9c  9B) sin 9B] =A.B+A.C. We used the identity for cos (a + {3)from Appendix. A. . Solution
(a) The angle between A and B is 60° + 10° = 70°, so AB = (10)(4) cos 70° = 13.7 (Note: units are ignored in this problem.) (b).In terms of components, C D = (5.6)(1.9) ~ (3.1)(7.2) = 11.7. (c) Express B in components (4 cos 1000ff.4 sin 1000j), or C by magnitude and direction (6.40 at 29.0° below the :rraxis). In either case, BC=(4cos1000)(5.6) + (4 sin 100°)(3,1) = 4(6.4) cos 129° = 16.1. Problem
12. (a) Find the scalar products i. i, j. j and k. k. " (b) Find the scalar products i. j, j. k, k. i. (c) Use the distributive law to multiply out the scalar product of two arbitrary vectors A=Azi+Ayj+Azk and B=Bzi+Byj+Bzk. Then use the results of parts (a) and (b) to verify Equation 74. Problem
15. (a) Find the scalar product of the vectors ai + bj and bi  aj, and (b) determine the angle between them. (Here a and b are arbitrary conStants.) Solution
(a) (ai+bj)(biaj)=ab+b(a) =0. (b) The vectors are perpendicular, so 9 = 90°. Solution
(a) The dot product of any vector with itself equals its magnitude squared, A. A = A 2 cos 0° = A 2• For any unit vector, fi.. fi.= 1. (b) If two vectors are perpendicular, their dot product is zero, A B = 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) AB=(Azi+Ayj+Azk)  (Bzi + Byj + Bzk) = AzBzi. i + AzByi j + AzBzi k + AyBJ i + AyBy~' j_+AyBzj k + AzBzk j + AzByk j + AzBzk k AzBz,+ AyBy AzBz. (We also used the commutative law, i. j = j i = 0, etc.) Problem
16. Rework Example 75 using Equation 73 instead of Equation 74. Solution
F has magnitude v'4902 + 2302 N and direction 9 = tan1(230 490 = 25.1° above the xaxis; I~rl = 1.7 + 1;90 mat 47.0° above the xaxis. Therefore, F ~r= (541 N)(2.60 m) cos(47.00  25.1°) = 1.30 kJ. = + Problem
13. One vector has magnitude 15 units, and another 6.5 units. Find their scalar product if the angle between them is (a) 27° and (b) 78°. Problem
17. Use Equations 73 and 74 to show that the angle \ between the vectors A=azi+ayj, and B=bzi+ byj is Solution
(a) AB=(15u)(6.5u)cos27° = 86.9u2• (b) A.B= (15u)(6.5u) cos 78° = 20.3u2• (Note: We used u as the symbol for the unspecified units.) Solution
AB=azbz + ayby = Ja~ +a~Jb~ + b~cosB, from which the desired expression for () follows directly. 110 CHAPTER 7 Problem
18. Find the angles between all three pairs of the vectors A= 3i+2.i, B::: i+6.i, C=7i2j. Problem
21. A force F=67i+23.i+55k N is applied to a body as it moves in a straight line from rl = 16i+31j to r2 =2li+ lOj+ 14k m. How much work is done by the force? Solution
The result of the previous problem gives the angle between any two vectors as 8AB = cosl(A. B/AB). (We are using 0AB for the magnitude of the angle between A and H, which by definition is in the range 0° to 180°.) Therefore: O=cosl
AB Solution
'iW=F.~r=F. (r2  rl) = [67(2116) :!31) + 55(14)] N.m = 622 J. + 23(10  'Problem
22. A rope pulls a box a horizontal distance of 23 m, as shown in Fig. 726. If the rope tension is 120 N, and if the rope does 2500 J of work on the box, what angle does it make with the horizontal? [. 3(1)+2(6) v'32 +22/(1)2 =6580 +62 ., o
AC. = cosl [ 3(7) + 2( 2) ] = 49.6° and v'I3/72 + (2)2 ' . 8 1[(1)7+6(2)]_1150 Be  cos v'37 v'53 Solution
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
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. 23~=Ax
FIGURE 726 Problem 22 Solution. Solution
The force is constant, and r is the displacement from the origin, so the work done by F is W = F • r = [(1.8)(56) + (2.2)(31)] N . m = 169 J. Section 73: Problem A Varying Force Problem
20. A force F = 14i + llj N acts on an object. Find the work done by the force if the object moves from the origin to the point (a) 28i + 22j m and (b) 22i  28j m. 23. Find the total work done by the force shown in Fig. 727 as the object on which it acts moves (a) from x = 0 to x = 3 kill; (b) from x = 3 kill to x=4km.
50 ....... 40 ~ 30 ..• Solution
The work done by a constant force F acting over a displacement r from the origin is W = F • r. For the given vectors, the scalar product can be calculated from Equation 74, Wa = (14x28 + llx22) J = 634 J for part (a), or Wb = (14x22 llx28) J = 0 for 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 part (b), since the product of the slopes is minus one ((Fy/Fx)(Y/x) =: 1). Then Equation 73 gives Wa = 2(142 + 112) J and Wb = O. ~20
10 1 2 3 4 Distance (Ian) FIGURE 727 Problem 23. Solution
The work done by a onedimensional force is (Equation 78) W = F(x) dx. From Fig. 727, F(x) is a linear function in the two intervals specified: 1:; F(x) = (40 N/3 km)x, 40 N  (40 N/km) x { (x  3 km), for 0 ::; x ::;3 km. for 3 km ::; x ::;4 km. CHAPTER7
(Use the slope/int~rcept. equation for a straight line, + b, to verify this.) Therefore: 3km (a) 111 Solution
(a) W =
2 y = mx WO3 = 1 G~~)
(40 N) I(4 krn)xkrn 20 kJ. 1:: F dx
= = fgmax2dx = la Ix31~m = x dx = (40 N) 3.krn (b) W3~ =
= h:~(4~:)(4 (3 km)2 = 60 kJ 2 ' km  x)dx
4km ~ 21 2 3km = (Of course, the triangular areas under the force vs distance curve could have been calculated in one's head; however, it is instructive to understand the general method for evaluating Equation 78.) 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 !::J.Xi= 2 m. Then W::::l (5 N/m2)(12 + 32 + 52) x m2(2 m) 350 J. The percent error is only 8 = 100(360  350)/360 = 2.78%. (c) Now, Xi = 0.5, 1.5, 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 + 52 + 72 + 92 + 112)(1 m) = 357.5 J, and 8 = 100(2.5)..;2 360 = 0.694%. (d) W::::l E~:l F(Xi)!::J.Xi (5 N/m )x 2 +32 + ... + 232)(0.5 m) (0.25 m)2(1 359.375 J with 8 = 0.174%. (The direct calculation of the sum is tedious, but we can use the formula for the sum of the squares of the first n numbers, namely E~ k2 = ~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
24. Findthe total work done by the force shown in. Fig. 728, as the object on which it aCts moves : from x = 0 to x = 5.0 m. 2300.) Problem
26. A spring has spring constant k = 200 N/m. How much work does it take to stretch the spring (a) 10 em from equilibrium and (b) from 10 cm to 20 em from equilibrium? Solution
The work is the area under the graph between x = 0 and 5 m, as shown. The seven squares and four triangles represent 10.5 J of work. Solution
(a) W = ~kx2 (from Equation 710) = ~(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 Distance (m) 5 6 FIGURE 728 Problem 24 Solution. 27. A certain amount of work is required to stretch spring A a certain distance. Twice as much work is required to stretch spring B half that distance. Compare the spring constants of the two springs. Problem
25. A force F acts in the x direction, its magnitude given by F = ax2, where x is in meters, and a is . 2 exactly 5 N/m . (a) Find an exact value for the work done by this force as it acts on a particle moving from x = 0 to x 6 m. Now find approximate values for the .work by dividing the area under the force curve into rectangles of width (b) !::J.x= 2 m; (c) !::J.x= 1 m; (d) !::J.x= m with height equal to the magnitude of the force in the center of the interval. Calculate the percent error in each c.ase. Solution
We are given
!kBX2 = 2WA = kAX2• WA = !kAx2 and WB = B(x/2)2 Therefore kB = 8kA• !k =
. Problem
28.. On graph paper, draw an accurate forcedistance curve for the force of Example 78, and obtain a ..solution to that example by determining graphically the area under the curve. = ! Solution
We can superimpose a grid on Fig. 716 and count rectangles. The area represented by one rectangle is 112 CHAPTER 7 1 kJ, and there are about six and one half such rectangles under the curve between x = 0 and 10 m. Thus, W ~ 6.5 kJ.
1500 Solution
The work is the area under a cosine curve, which the student may plot if so required. The force reverses direction when x = !1rXo ~ 20.4 m. The "area" 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 by this force acting on an object as it moves (a) from x = 0 to x = 3 m; (b) from x = 3 m to x = 6 m; (c) from 6 m to 9 m. 2 4 6 B Position, % (Ill) 10 FIGURE 716 Problem 28 Solution. Solution Problem
29. A force F. acts in the x direction; its x component is given by F = Fo cos(x/xo), where Fo = 51 N and Xo = 13 m. Calculate the work done by this . force acting on an object as it movesfrom x = 0 to x = 37 m. Hint: Consult Appendix A for the integral of the cosine function and treat the argument of the cosine as a quantity in radians. From Equations 78 and 79:
W
Xl+X2 _lx2
Xl ax 1/2dx _ ax 32 / 1 x2
Xl  (3/2)
xl'  '3 _ (2a) (x 3/2 _
2 3/2) Solution
From Equation 78: W = Therefore, (a) WO+3 = (2/3)(9.5 N/ml/2)(3 m)3/2 32.9 J, (b) W3+6 = (6.33 N.m1/2)[(6 m)3/2 (3 m)3/2] = 60.2 J, and (c) W6+9 = 77.9 J. = Problem
Xo 1
o 37m Fo cos =dx = Fo I xo sin = 1 xO
o 37m = (51 N)(13 m) sin(37/13) = 193 J.
Problem
30. Work the preceding problem graphically by making an accurate plot of the forceversusdistance curve on graph paper, and determining the area (in units of work) under the curve. 32. A force given by F = b/..jXacts in the x direction, where b is a constant with the units N.ml/2. Show that even though the force becomes arbitrarily 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. Solution W = bXI/2dx = 12bxl/21:: = 2b(y'x2  y'xl), which is finite for Xl + O. In fact, W = 2bJX2, for Xl + O. 1:: Problem
33. The force exerted by a rubber band is given approximately by x F=Fo eo +x [ e;; (iO+x)2 e5] ' F;a>5 (X/x,,)
Problem 30 Solution. where eo is the unstretched length, X the stretch, . and Fo is a constant (although Fo varies with temperature). Find the work needed to stretch the rubber band a distance x. CHAPTER7 113 w = t Fo [£0 £0 + 10
1 ( , X' _ £~ ] dx' (£0 + x')2
Z = Fo 1 £0 lox = X'2) £~ I + 2" + £0 + x' 0 2 FO(X+ 2£0 + ~ x £o+x (0)' parallel to dr along the path, then F.dr=Fldrl, and W =F Idrl, where the scalar integral is now just the length of the semicircular path. Thus W = 1rRF. (Note that the symbol dr is a displacement along the path, so Idrl is an element of path length, not the differential of the radius. Where confusion might arise, one can use dl for path element, as in Chapters 25, 30, and 31.) I! (x' is a dummy variable) (I) +~
(I) Section 74: Force and Work in Three Dimensions Problem
34. A car drives 3.1 kIn southward, propelled by the 6.3kN force of friction between its tires and the road. It then turns eastward and goes another 1.8 kIn; during this stretch the frictional force is :' 5.1 kN. Finally, the car turns toward the southeast' 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 ' the direction of the car's motion. A) It ,/ ~ ~I I. (b) .••.•.•.• I I " ~
(b) .
I (c) A.
FIGURE • {  l_   ...•. X ~
B 729 Problem 35 Solution. Problem
36. A cylindrical log of radius R lies half buried in the ground, as shown in Fig. 730. An ant of mass m climbs to the top of the log. Show that the work done by gravity on the ant is mgR. Solution
Since the frictional force is in the direction of motion on each segment, this problem reduces to three onedimensional 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. 729. It is subject to a force of constant magnitude F. Find the work done by the force (a) if the force always points upward in Fig. 729, (b) if the force always points to the right in Fig. 729, and (c),if the force always points in the . direction of the particle's motion. FIGURE 730 Problem 36. Solution
If we let dr=dxi+dyj, as in Example 79 (wherei is horizontal to the right and j is vertical upward) then Wg = f~Fg .dr= f~(mgj) • (dxi+dyj) = my dy = mg(Y2  Yl) = mgR. (The difference in height going from the ground to the top of a halfburied log is just the radius.) Another way of obtaining this result is to use reasoning similar to that in the solution to Problem 35. The force of gravity is constant and can be taken outside the integral, Wg = I~ 9 • dr = F9 • Ii dr.The integral left is the total F displacement, dr = Rj  Hi (if we take origin at the center of the log's croSssection), so Wg = (mgj) .(Rj  Ri) = mgR. Finally, one could use Solution
(a) and (b) If the force is a constant vector (in magnitude and direction) it may be factored out from under the integral in the work (which is the limit of a sum) to yield W = F • dr=F . dr. The remaining integral is the sum of the displacements around a semicircle, which is just the total vector displacement along the diameter, from A to Bin Fig. 729. If we introduce 1ry coordinates to the right and upward, respectively, with origin at the center of the semicircle, then If dr=AB= 2Ri. For F=F1, . W = Fj.2Ri="O, and for F=Fi, W = Fi.2Ri= 2RF. (c) If the force is constant in magnitude and f:~ I! I! !i
I' 'i Ii 114 CHAPTER 7 the work becomes: W = .
2 Equation 73 for the dot product, as shown: Wg = 1
2 F 9 • dr =  11
2 r2 mg cos 0 Idrl = r/ 10 mgcosO.RdO=mgRlsinOI~/ =mgR. 1 1 = +1 = +1 =
(3'0) F .i dx + /(3.6) F .j dy
6 (0,0)
3 (3,0) = Fz(Y 0) dx Fy(x = 3) dy
= 90 J. 6 0 d dy (15 N)(6 m) (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(lcosrjJ). Problem 36 Solution. Solution Problem
37. A particle of mass m moves from the origin to the point x = 3 m, y = 6 m along the curve y = ax2  bx, where a = 2 mI and b =4. It is subject to a force F = cxyi + dj where c = 10 Njm2, and d = 15 N. Calculate the work done by the force. The path is a circular arc (of radius i and differential arc length Idrl = idO) so that only the tangential (or parallel) components of any forces acting do work on the swing; the radial (or perpendicular) components do no work since the dot product with the path element is Zero. Thus, the tension in the chains and the radial components of gravity or the applied force do no work. If you pull slowly, so that the tangential acceleration is zero, then I'll = mgsinO, and the work you do is Solution
Since the equation for the curve gives y as a function of x, we can eliminate y and dy in the line integral for the work: W = = lr~ 1
r) w= 10 rei> F.dr= 10 rei>, FJildrl = 10 rei> mgsinO.£d(} F . dr =
2 {(3.6) 1(0,0) (Fxdx + Fy dy)
 b)) dx  bx )
3 10 = mgi 1 cos Ol~ = mgi(l  cos rjJ). 3 [cx(ax  bx) + d(2ax
2 + d (x2aT (Since forces perpendicular to the path of the swing do no work, all the work you do, for zero tangential 3 x4 'X = c ( aT  b"3 )
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 xaxis from the origin to the point (3,0), then parallel to the yaxis 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 acceleration, is done against gravity, W =  Wg• But my Ay (as in Example 79), where Ay is the ch~ge in height of the swing. In this case, Ay £ _ £cosljJ, which gives the same Was above.) 115 w: = Solution
We want !mcvb = !mTvf, so Vc (20 km/h) /(3.2xl04 kg)/(950 kg) 116 km/h. = = vT/mT/mc = = 5.80VT = Section 75: Problem Kinetic Energy Problem
44. A 60kg skateboarder comes over the top of a hill at 5.0 mis, and reaches 10 m/s at the bottom of the hill. Find the total work done on the skateboarder between the top and bottom of the hill. 40. A 2.4x 105 kg jet is cruising at 900 km/h. What is the kinetic energy relative to the ground? A 65kg passenger strolls down the aisle at 3.1 km/h. What is the passenger's kinetic energy in the reference frame of (a) the plane and (b) the ground? Solution
(a) Kjet = !(2.4x105 kg)(900 m/3.6 S)2 7.50 GJ (see, Table 11). (b) Relative to the plane,K~ass = !(65 kg), (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 passenger is 900:1: 3.1 lan/h, depending on his or her direction down the aisle. Therefore K pass !(65 kg)(900:l: 3.1)2(1 m/3.6 s)2 .;" 2.05 MJ or 2.02 MJ. Solution = The workenergy theorem, Equation 716, gives Wnet = AK = ~m(v~  v?) = !(60 kg) x (102  52)(m/s) 2.25 kJ. . = Problem
45. Two unknown elementary particles pass through a detection chamber. If they have the same kinetic energy and their mass ratio is 4 : 1, what is the ratio of their speeds? = Problem
41. Electrons in a color TV tube are accelerated to 25% of the speed of light. How much work does the TV tube do on each electron? (At this speed, relativity introduces small but measurable corrections; here you neglect these effects.) See inside front cover for the electron mass. Solution
If
ml == 4m2 and Kl == K2, then Vl == /2Kl/ml = /m2vUml == v2/m2/ml == !V2; a mass ratio of 4 : 1 corresponds to a speed ratio of 1 : 2 if the kinetic energy is the same. Problem
46. You do 8.5 J of work to stretch a spring of spring constant k == 190 N/m, starting with the spring \ unstretched. How far does the spring stretch? Solution
From ,the workenergy theorem, Wnet == AK == !me(0.25 C)2 0.5(9.11xlO3l kg)(0.25x3x 108 m/s)2 = 2.56xlO15 J = 16.0 keV. = Solution
The work done on a spring in stretching it a distance from its unstretched length is (Equation 710 W = !kx2. Thus x = /2W/k == 2(8.5 J)/(190 N/m) == 29.9 cm. (See also Example 77(a).) Problem
42. In a cyclotron used to produce radioactive isotopes for mediCal research, deuterium nuclei (mass 3.3x 1027 kg) are given kinetic energies of 8.8xlOl3J. What is their speed? Compare with the speed of light. x Problem
47. ~fter a tornado, a O.5Gg drinking straw was found embedded 4.5 em in a tree. Subsequent ' 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? Solution
v
== J2K m = [2(8.8XlO l3 J)]1/2 3.3x 1O27kg = 2.31x107 m/s = 0.077c Problem
43. At what speed must a 950kg subcompact car be movingto have the same kinetic energy as a 3.2xI04_kg truck going 20 km/h? Solution
Since the stopping force (70 N) is so much larger than the weight of the straw (0.0049 N), we may assume that the net work done is essentially that done by just the stopping force, and use the workenergy theorem, 116 CHAPTER 7 Wnet = I:1K.The force is opposite to the displacement, so F I:1r = 0  !mv2, or v = v2Fl:1r/m = V2(70 N)(0.045 m)/0.5xlO3 kg = 112 m/s ( 250 mifh). Problem
48. You drop a 15Dgbaseball from a sixthstory window 16 m above the ground. What are (a) its kinetic energy and (b) its speed when it hits the ground? Neglect air resistance. I:1K = O.The work done by gravity is Wg = mgx (72 cm), and that done by the stopping force is WF = F(2 cm). Therefore, mg(72 cm) F(2 cm) = 0, or F = (8 kg)(9.8 mji)(72/2) = 2.82 kN = 635 lbs. Problem
50. From what height would you have to drop a car for its impact to be equivalent to a collision at 20 mph? Solution
(b) The speed of the baseball (magnitude of the velocity) followsfrom Equation 211, the initial conditions (with y = 0 at ground level) and the neglect of air resistance: v = V2g(y  Yo) = J2(9.8 m/s2)(16 m) = 17.7 m/s. (a) The kinetic energy is K = ~mv2 = !(0.150 kg)(17.7 m/s)2 = 23.5 J. Alternatively, K = ~m(2gyo) = mgyo, and v= v2K/m. Solution
"Equivalent" means the same kinetic energy is lost by the car as in the collision. Therefore, the impact speed is 20 mi/h = J29Ti, or h = (20x1.609 m/3.6 s)2+ 2(9.8 m/s2) = 4.08 m = 13.4 ft; Problem
51. Catapults run by highpressure steam from the ship's nuclear reactor are used on the aircraft carrier Enterprise to launch jet aircraft to takeoff speed in only 76 m of deck space. A catapult exerts a 1.1x106 N force on a 3.3x104 kg aircraft. What are (a) the kinetic energy and (b) the speed of the aircraft as it leaves the catapult? (c) How long does the catapulting operation take? (d) What is the acceleration of the aircraft? Problem
49. A hospital patient's leg slipped off the stretcher and his heel hit the concrete floor. As a physicist, you are called to testify about this accident. You estimate that the foot and leg had an effective mass of 8 kg, that they dropped freely a distance of 70 cm, and that the stopping distance was 2 cm. What force can you claim the floor exerted on the foot? Give your answer in pounds for the jury's sake. Solution
(a) If we assume that the carrier deck and catapult force are horizontal, then the catapult force is the net force acting on the aircraft. The workenergy theorem (Equation 716) gives I:1K = K  0 = Wnet = Fnetl:1x = (l.lx106 N)(76 m) = 83.6 MJ, for an aircraft starting from rest. b From Equation 715, v = V2K/m = 2(83.6 MJ)/(3.3x104 kg) = 71.2 m/s = 256 kID/h. (c) From Equation 29, t = 2(x  xo)/(vo + v) = 2(76 m)/(O + 71.2 m/s) = 2.14 s. (d) The acceleration can be calculated from Fnetlm = 33.3 m/s2, or from Equation 27, a = vito In fact, v/t = v/(2 I:1x/v) = v2/2 I:1x = (2K/m)/(2I:1x) =
Fnet/m. Solution
This problem can be solved with the use of Newton's second law and kinematics (see Problem 567, for example), but alternatively, we may apply the workenergy theorem: Wnet = Wg + W F = I:1K. The leg starts from rest at point A and stops at point B, so Section 76: Problem \ Power 52. A horse plows a 20DmIongfurrow in 5.0 min, exerting a force of 750 N. What is its power output; measured in watts and in horsepower? Solution
Problem 49 Solution. If the force is assumed parallel to the displacement, 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
53. A typical car battery stores about 1 kWh of energy. What is its power output if it is drained completely in (a) 1 minutej (b) 1 hourj (c) 1 day? launch? (b) If the driver makes one launch every 30 min, what is its average power consumption? Solution
(a) The work done by the driver during a launch is equal to the change in kinetic energy of the package, so the power output is P = Wit = D.Klt = ~(103 kg)(2 km/s)2/55 s= 36.4 MW. (b) The same work (one launch) spread over 30 min yields an average power of Pay = D. WI D.t = 2 x 109 J ..;30x60 s = 1.11 MW. (Note: we are neglecting any work done by lunar gravity during the launch, which would be a small c~rrection to W.) Solution
The average power (Equation 717) is (a) P = (1 kWh)/(1 h/60) = 60 kWj (b) 1 kWh/1 h = 1 kWj and (c) 1 kWh/24 h = 41.7 W, Problem
54. A sprinter completes a 1ODmdash in 10.6 s, doing 22.4kJ of work. What is her average power output? Problem
58. A75kglongjumper takes 3.1 s to reach a prejump speed of 10 m/s. What is his power output? Solution
From Equation 717, P = D.W/D.t = 22.4 kJ/10.6 s = 2.11 kW. (The distance of the dash is not relevant here.) Problem
55. How much work can a 3.5hp lawnmower engine do in 1 h? Solution P = Wit = D.K/t = !(75 kg)(l0 m/s)2/3.1 1.21 kW ~ 1.6 hp. s= Problem
59. Estimate your power output as you do deep knee bends at the rate of one per second. Solution
Working at constant power output, Equation 717 gives the total work (energy output) as D.W = P D.t = (3.5 hp) (746 W/hp)(3600 s) = 9.40 MJ. (Note the change to appropriate SI units.) Solution
The work done against gravity in raising or lowering a weight through a height, h, has magnitude mgh. The body begins and ends each deep knee bend at rest (D.K = 0), so the muscles do a total work (down and up) of 2mgh for each complete repetition. If we assume that the lower extremities comprise 35% of the 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 (9.8 m/s ) = 480 N. We guess that h is somewhat greater than 25% of the body height, or about 45 em, so the muscle power output for one repetition per second is about 2(480 N)(0.45 m)/s =430 W. Problem
56. Water drops over 49mhigh Niagara Falls at the rate of 6.0 x 106 kg/so If all the energy of the falling water could be harnessed by a hydroelectric power,: plant, what would be the plant's power output? Solution
The force of gravity does work dWg = (dm)g D.y on each mass dm of water going over the falls. If we 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, is given, the available power can be found: P = dWg/dt = (dm/dt)g D.y = (6x106 kg/s)x 2 (9.8 m/s )( 49 m) = 2.88 GW. Problem
60. At what rate can a onehalf horsepower well pump deliver water to a tank 60 m above the water level in the well? Give your answer in kg/s and gal/min.
\ Problem
57. A "mass driver" is designed to launch raw material mined on the moon to a factory in lunar orbit. The driver can accelerate a 100Dkg package to 2.0km/s Uust under lunar escape speed) in 55 S. (a) What is its power output during a Solution
If D.m/ D.t is the mass of water pumped per second, the work done (lifting D.m against gravity at constant speed) per second is P = D.W/D.t = (D.m/D.t)gh. Therefore, D.m/D.t = P/gh = (0.5 hp)(746 W/hp)+ (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 1O3m3/gal) = 3.786 kg/gal, so 6.m/6.t (0.634 kg/s)(60 s/ min)/(3.786 kg/gal) = 10.1 gal/min. Problem
65. By measuring oxygen uptake, sports physiologists have found that the power output of longdistance runners is given approximately by P = m(lw  e), where m and v are the runner's mass and speed, respectively, and where band c are constants given by b = 4.27 J/kg.m and e = 1.83 W /kg. Determine the work done by a 54kg runner who runs a 100km race at a speed of 5.2 m/s. = Problem
61. In midday sunshine, solar energy strikes Earth at 2 the rate of about 1 kW /m • How long would it take a perfectly efficient solar collector of 15.m 2 area to collect 40 kWh of energy? (This is roughly the en.ergy content of a gallon of gasoline.) Solution
The runner's average power output is P = (54 kg) x [(4.27 J/kg.m)(5.2 m/s) 1.83 W/kg) = 1.10 kW. Over the race time, 10 km/(5.2 m/s) = 1.92x 103 s, the runner's work output is 6.W = P 6.t = 2.12 MJ 0.588 kWh. Solution
The average power received by the collector is (1 kW /m2)(15 m2) = 15kW, so it would take 6.t = 6.W/P = 40 kWh/15 kW = 2.67 h to collect the required energy. (The average intensity of sunlight is discussed in more detail in Chapter 34.) = Problem
66. A 65kg runner running at Vo = 4.8 m/s accelerates to 6.1 m/s over a 25s interval. (a) By writing v = Vo + at, where a is the runner's acceleration, use the formula in the previous problem to express the runner's power output as a function of time. (b) How much work does the runner do during the acceleration period? Problem
62. It takes about 20 kJ to melt an ice cube. A typical microwave oven produces 625 W of microwave power. How long will it take to melt the ice cube in this oven? Solution
From Equation 717,6.t 32.0 s. = 6.W/P = 20 kJ/625 W = Solution
(a) P(t) = m(bvo + bat  e). (b) W = f~' P(t) dt = f~,=25S m(lwo + bat  e) dt = m(lwotl + !bat~  ett} = mtl[!b(vo + Vl) c) = (65 kg)(25 s)[!(4.27 J/kg.m) (4.8 Problem
63. The rate at which the United States imports oil, expressed in terms of the energy content of the imported oil, is nearly 700 GW. Using the "Energy Content of Fuels" table in Appendix C, convert this figure to gallons per day. + 6.1)(m/s)  1.83 W Ikg) = 34.8 kJ. Problem
67. A 1400kg car ascends a mountain road at a steady 60 km/h. The force of air resistance on the car is 450 N. If the car's engine supplies energy to the drive wheels at the rate of 38 kW, what is the slope angle of the road? Solution
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 430 million gallons per day. Solution
At constant velocity, there is no change in kinetic energy, so the net work done on the car is zero. Therefore, the power supplied by the engine equals the power expended against gravity and air resistance. The latter can be found from Equation 721, since gravity makes an. angle of () + 90° with the velocity (where () is the slope angle to the horizontal), while air resistance makes an angle of 180° to the velocity. Then 38 kW = Fg VFair v = mgvcos«(} + 90°) Fair V cos(1800) = mgv sin () + Fair v, or () = sin 1 x [«38 kW 160 km/h)  450 N)/(1400 kgx 9.8 m/s2») = 7.67°. (See Example 714 and use care with SI units and prefixes.) Problem
64. Which consumes more energy, a 1.2kW hair dryer used for 10 min or a 7W night light left on for 24 h? Solution
The energy consumption of each device is 6.W = P6.t; (1.2 kW)(1 h/6) = 0.2 kWh for the hair dryer and slightly less, (7 W)(24 h) = 0.168 kWh, for the night light: CHAPTER 7 119 Problem
68. A machine does work at a rate given by P = where c = 18 W /S2, and t is time. Find ~n expression for the work done by the machine between t = 10 s and t = 20 s. Problem ct ,
.
2 • 72. A force pointing in the x direction is given by F = ax3/2, where a == 0.75 N/m3/2. Find the work done by this force as it acts on an object moving from x = 0 to x = 14 m. Solution
Equation 718 implies Solution W= 1 205 1205 dt Pdt ct
=
2 lOs W = 1
o 14m ax3/2 dx = aX 5/2/l4m 5 0 (2) 105 = ~ (18 ~) =42 kJ. [(20 s)3  (10 s)3] = ~ (0 75~) 5 . m3/2 (14 m)5/2 = 220 J Problem Paired Problems Problem
69. You apply a 470N force to push a stalled car at a. 17° angle to its direction of motion, doing 860 J of work in the process. How far do you push the car? 73. Two vectors have equal magnitude, and their scalar product is onethird of the square of their magnitude. Find the angle between them. Solution
We are given that A2 = B2 = 3A.B Therefore () = cosl(I/3) = 70.5°. = 3ABcos(). Solution
Equation 72 or 75 gives the work done by the applied force; hence !:i.r = W/Fcos() = 860 J/(470 Nx cos 17°) = 1.91 m. Problem
74. Vector A has magnitude A, vector B has magnitude 2A, and A. B = A 2• Find the angle between A and B. Problem
70. A tractor tows a jumbo jet from its airport gate, doing 8.7 MJ of work. The link from the plane to the tractor makes a 22° angle with the direction of the plane's motion, and the tension in the link is 4.1xl05 N. How far does the tractor move the plane? Solution
A2 = A. B = AB cos () = 2A2 cos ()j therefore () =
cosl(1/2) = 60°. Problem
75. A 460kg piano is pushed at constant speed up a ramp, raising it a vertical distance of 1.9 m (see Fig. 732). If the coefficient of friction between Solution
As in the previous problem, !: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 onedimensional given by Equation 78: situation is W = rO(Fo)x 10 dx = (Fo)
Xo x~ = ~Foxo.
2 2
FIGURE 732 Problem Xo 75. 120 CHAPTER 7 ramp and piano is 0.62, find the work done by the agent pushing the piano if the ramp angle is (a) 15° and (b) 30°. Assume the force is applied ._ parallel to the ramp. Problem
77. (a) How much power is needed to push a 95kg chest at 0.62 m/s along a horizontal floor where the coefficient of friction is O. 8? (b) How much 7 work is done in pushing the chest 11 m? Solution
The usual relevant forces on an object pushed up an " incline of length l = hi sin 9 by an applied force parallel to the slope are shown in the sketch. At constant velocity, the acceleration is zero, so the parallel and perpendicular components of Newton's second law, together with the empirical relation for kinetic friction, give N = mgcos9, fk = ILkN, and Fapp = mg sin9 + fk = mg(sin 9 + ILkcos 9). Thus, the work done by the applied force is Wapp = Fappl = Fapphl sin 9 = mgh(1 + ILkcot 9), where h is the  vertical rise. (a) Evaluating the above expression using the data supplied, we find Wapp = (460 kg) x (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 calculatiion,we find Wapp = 17.8 kJ. (The work done against gravity is the same in parts (a) and (b) since h is the same, but the work done against friction is greater in (a) because the incline is longer and the normal force is greater; however Fapp is less.) Solution
If you push parallel to the floor at constant velocity, the normal force on the chest equals its weight, N = mg, and the applied force equals the frictional force, Fapp =!k = ILkN = ILkmg. (a) The power required is (Equation 721) Papp = Fappv = (0.78)x (95 kg)(9.8m/s2)(0.62 m/s) = 450 W, or about 0.6 hp. (b) The work done by the applied force acting over a displacement tlx = 11 In is Wapp = Fapp.6.x = Papp(.6.xlv), where t = .6.xlv is the time over which the power is applied. Using either expression, we find Wapp = 7.99 kJ. Problem
78. You mix flour into a thick bread dough, exerting a 45N force on the stirring spoon. If you move the spoon at 0.29 mis, (a) what power do you supply? (b) How much work do you do if you stir for 1.0 min? i
h Solution
(a) Provided the stirring force is applied always parallel to the velocity of the spoon, Papp = Fappv = (45 N)(0.29 m/s) = 13.1 W. (b) W~pp = Pappt = (13.1 W)(60 s) = 783 J. J
Problem 75 Solution. Supplementary Problems Problem
79. The power output of a machine of mass m increases linearly with time, according to the formula P = bt, where b is a constant. (a) Find an expression for the work done between t = 0 and some arbitrary time t. (b) Suppose the machine is initially at rest and all the work it supplies goes into increasing its own speed. Use the workenergy theorem to show that the speed increases linearly 'with time, and find an expression for the acceleration. Problem
76. You have to do 2.2 kJ of work to push a 78kg trunk 3.1 m along a slope inclined upward at 22°, pushing parallel to the slope. What is the coefficientof friction between trunk and slope. Solution Using the equation derived for Wapp in the previous problem, one can solve for the coefficient of friction: ILk = Solution
(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, then !bt2 = ~mv2, since the machine starts from rest. Thus v = Jblm t and a = dvldt = Jblm. (v is the speed and a is the tangential acceleration along the path of the machine.) (W app mgh  1) tan 9 _ ( 2.2kJ 1)  (78 kg)(9.8 m/s2)(3.1 m) sin 22° x tan 22° = 0.60. CHAPTER 7 121 em
ou're trying to decide whether to buy an ergyefficient,225W refrigerator for $1150 or a dard, 425W model for $850. The standard ..odel will run 20% of the time, while better ulation means the energyefficient model will irun 11% of the time. If electricity costs 9.5t/kWh, how long would you have to own the energy','efficient odel to make up the difference in cost? m 'Neglect interest you might earn on your money. uti on e price differential is equal to the difference in cost of energy over a period of time t, then (1150 850) = (Pstdx20%t  PeffX l1%t) x $0.095/kWh). Solving for t, we find t = (300)x [(0.425xO.2 0.225xO.11)(0.095)]1 h = 5.24x104 h = 2.18x103 d ~ 6.0 y. Solution
If we ignore the mass of the spring, the spring tension increases from 0 to mg in the interval before the mass leaves the ground. The work done against the spring force is W = Fsprdx, as in Example 77. Since we know Fspr as a function of X, Fspr = kx, we can find the appropriate limits for the interval Xl = 0 to X2 = mg Ik and integrate over x. The result is W = {:9/k kx dx = !k(mglk)2 = m2g2/2k. (Alternatively, we can use Fspr = kx to eliminate X, and integrate over Fspr as a variable. This gives dFspr = k dx and W = 1;"9 Fspr(dFsprlk) = Hmg)2Ik.) I:: 'Jl :"] Problem
83. Figure 733 shows the power a baseball bat delivers to the ball, as a function of time. Use graphical integration to determine the total work the bat does on the ball. .Problem
81. The percapita energy consumption rate plotted in Fig. 721 can be approximated by the expression P = Po +at + bt2 + et3, where Po = 4.4 kW, a = 5.57xl02 kW Iy, b = 3.84xlO3 kW ly2, C = 2.79xlO5 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 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 kilowattyears. FIGURE 733 Problem 83. Solution
The work done on the ball, W = 1 Pdt, is the area under the graph of power versus time, where each small rectangle in Figure 733 has an "area" of (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. Solution
The energy used between times h and
2 t2 is W = {t p dt
1tl = I
tl h (Po + at  tn + bt2 + ct3)dt + ib(t~
 t~) = PO(t2  h) + !a(t~ +ic(t~ t1), (a) Using tl = 40 y, t2 = 50 y, and the given coefficients,we find the energy used in the 1940s to be W = 71.3 kW.y. (b) A similar calculation for the 1960sgives W = 93.3 kW.y. Problem
84. A machine delivers power at a decreasing rate P = PotV(t + to)2, where Po and to are constants. The machine starts at t = 0 and runs forever. Show that it nevertheless does only a finite amount of work, equal to Poto. Problem
82. A spring of spring constant k is attached to a mass 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 done on the spring before the mass begins to leave the ground. Solution
From Equation 720, W~ 10 r~)Pot5 dt (t + to)2 =_ I(t+ to) /00 Pot5
0 = Poto. Problem
85. An unusual spring has the forcedistance curve shown in Fig. 734 and described by F = 100x2 for ...
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This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.
 Spring '08
 WORMER
 Physics, Energy, Power, Work

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