Solutions for chapter 6

Solutions for chapter 6

Questions:

2. No, because the force makes a 90 degree angle with the displacement.

5. No. Consider 2 cases, first where the dishes don't move at all (zero work done), and second, where the dishes move slightly in the direction of the pulled tablecloth. In this latter case the table cloth friction is exerting a force on the dishes in their direction of motion, so that W>0.

7. F1=k1*x1 F2=k2*x2

(a) If F1=F2 then k1*x1 = k2*x2, so that x2 = x1*(k1/k2)

W1 = 1/2 k1*x1^2

and W2 = 1/2 k2*x2^2 = 1/2 k2*(x1*(k1/k2))^2 = 1/2 k1*x1^2(k1/k2) = W1*(k1/k2) > W1,

because k1>k2

(b) If x1=x2 then W1=1/2 k1*x1^2, and W2 = 1/2k2*x2^2=1/2k2*x1^2.

Thus W1 > W2, (reverse of previous case) since k1>k2.

Problems

Sorry, I can't do Greek synbols in html, so I'll use the words theta, mu, sigma and Delta for those Greek letters

2. (a) W = F (cos theta) Delta x = 180N (cos 0) 6 m = 108 J

(b) W = F (cos theta) Delta x, where F = mg, and theta = 0 again, so W = (900 kg)(9.8 m/s^2)(6.0 m)= 5.29x10^4 J

10. Set up the free body diagram, and consider the x-component of each force, where the x-axis is directed up the incline, then sigma Fx = ma=0, since no acceleration. Call the man's force, F, and note that friction is directed up the incline if the piano is sliding down. From the wording of the problem, we can infer that the man needs to push up the incline also to keep the piano from sliding down.

Thus, sigma Fx = F + f - mg sin theta= 0, so that F = - f + mg sin theta . We need to find f here.

Use sigma Fy = ma=0, which gives N- mg cos theta = 0, so that N= mg costheta , and hence f = mu N = mu mg cos theta,

Substitution gives F = -f +mg sin theta= -mu mg cos theta + mg sin theta= (-0.4 cos 30 + sin 30)(280)(9.8) = 421.5 N

(b) W = F cos 180 Delta x = -(421.5)(4.3) = -1813 J (Man does negative work because the direction of his force is opposite to the displacement, and cos 180 = -1)

(d) W = F cos (90 - therta)Delta x = mg sin theta Delta x = (280)(9.8) (sin 30)(4.3) = +5900 J

(e) W net = +5900 - 4087 - 1813 = 0, which it had to be here since net F = 0 if v = constant.

13. When the force is varying, the work done is equal to the area under the F vs x curve.

(a) From x = 0 to x = 10, the resulting trapezoid can be thought of as two right triangles plus a rectangle, whose total area is 2800 N.m = 2800 J

(b) From x = 10 to x =15, the area is negative, and its value looks like around -700 N.m = -700 J, giving a total area from x = 0 to x = 15 of 2100 J.

18 (a) K = 1/2 mv^2, so v = sqrt(2K/m). Thus, if v à 2v, then K à sqrt(2) K = 1.41 K

(b) If v is doubled, Kà 4K

29. PE = 1/2 kx^2, so x = sqrt(2PE/k) = sqrt(2*25/440) = 0.337 m

39. (a) Use conservation of mechanical energy: PE1 + KE1 = PE2 +KE2, or mgy + 1/2mvo^2 = 0 + 1/2 mv^2,

giving v = sqrt(vo^2 + 2gy) = sqrt(5.0^2 + 2*9.8*3.0) = sqrt(83.8) = 9.15 m/s

(b) PE1 + KE1 = PE2 + KE2. For simplicity, assume that compression distance changes only the elastic PE of the trampoline, and neglect the change in gravitational PE.

0 + 1/2 mv^2 = 1/2 kx^2 + 0, so that x = sqrt(mv^2/k) = sqrt((75*9.15^2)/5.2x10^4) = 0.348 m

47. Use conservation of mechanical energy again: KE1 + PE1 = KE2 + PE2

so, 1/2 mv^2 + 0 = 0 + 1/2 kx^2. This gives for the max compression of the spring: x = sqrt(mv^2/k). To find the max force, use: F= kx = k sqrt(mv^2/k) = sqrt(kmv^2). Saying that we want this to be limited to 5g's means that F = 5mg, or that sqrt(kmv^2) = 5mg. Finally, solving for k gives k = 25 mg^2/v^2 = 25(1200)(9.8^2)/(27.8^2) = 3728 N/m (Note the need to convert v to m/s here)

58. p = W/t = mgy/t, so t = mgy/p = (285)(9.8)(16.0)/1750 = 25.5 sec