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Download mP = 1.67 x 10-27 kg, a = 3.6 x 1015 m/s2, v0 = 2.4 x 107 m/s, ∆x
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Physics 151 Key for Chapter 7 Homework 3. Givens: mP = 1.67 x 10-27 kg, a = 3.6 x 1015 m/s2, v0 = 2.4 x 107 m/s, ∆x = 3.5 cm. (a) From Table 2-1, we have v2 = v02 + 2a∆x. Thus, v = v02 + 2a∆x = (2.4 x 107 )2 + 2(3.6 x 1015)(0.035) m/s = 2.9 x 107 m/s. (b) The initial kinetic energy is Ki = mv02 /2 = (1.67 x 10-27 kg)(2.4 x 107 m/s)2 /2 = 4.8 x 10-13 J. The final kinetic energy is Kf = mv2/2 = (1.67 x 10-27 kg)(2.9 x 107 m/s)2/2 = 6.9 x 10-13 J. The change in kinetic energy is ∆K = (6.9 - 4.8) x 10-13 J = 2.1 x 10-13 J. 9. Givens: Figure 7-26, m = 3.0 kg, F is constant to right, 0.5 s intervals. Since this involves constant-acceleration motion, we can apply the equations of Table 2-1, such as x = v0t + at2/2 (where x0 = 0). We choose to analyze the third and fifth points (although any other pair of points could be used), obtaining 0.2 m = v0(1.0 s) + a(1.0 s)2 /2 -> 0.4 m = v0(2.0 s) + a(1.0 s)2 0.8 m = v0(2.0 s) + a(2.0 s)2 /2 Simultaneous solution of the equations leads to v0 = 0 and a = 0.40 m/s2 . We now have two ways to finish the problem. One is to compute force from F = ma and then obtain the work from Eqn. 7-7. The other is to find ∆K as a way of computing W (in accordance with Eqn. 7-10). In this latter approach, we find the velocity at t = 2.0 s from v = v0 + at (so v = 0.80 m/s). Thus, W = ∆K = (3.0 kg)(0.80 m/s)2/2 = 0.96 J. 15. Givens: F = 12.0 N (direction to be determined), d = (2.00i - 4.00j + 3.00k)m, (a) ∆K = 30.0 J, (b) ∆K = - 30.0 J. Using the work-kinetic energy theorem, we have ∆K = W = F• d = Fd cos(φ) In addition, d = (2.00)2 + (-4.00)2 + (3.00)2m = 5.39 m (a) ∆K = 30.0 J -> cos(φ ) = (30.0 J)/(12.0 N x 5.39 m) = 0.464 -> φ = 62.3O. (b) ∆K = - 30.0 J -> cos(φ ) = (- 30.0 J)/(12.0 N x 5.39 m) = - 0.464 -> φ = 118O. 17. Givens: m = 72 kg, d = 15 m upward, a = g/10 upward. (a) We use F to denote the upward force exerted by the cable on the astronaut and mg as the downward force of gravity on the astronaut. This could be shown on a simple free-body diagram similar to Figure 5-19. From Newton’s second law, Fnet = ma = F - mg = mg/10 -> F = 11mg/10. Since the force F and displacement d are in the same direction, the work is positive and the work done by the force F is just WF = Fd = 11 mgd/10 = 11(72 kg)(9.8 m/s2 )(15 m)/10 = 1.164 x 104 J which (with respect to significant figures) should be quoted as 1.2 x 104 J. (b) The force of gravity has magnitude mg and is opposite in direction to the displacement. Thus, using Eqn. 7-7, the work done by gravity is Wg = - mgd = - (72 kg)(9.8 m/s2)(15 m) = - 1.058 x 104 J, or - 1.1 x 104 J. (c) The total work done is W = (1.164 - 1.058) x 104 J = 1.06 x 103 J. Since the astronaut started from rest, the work-kinetic energy theorem tells us that this (which we round to 1.1 x 103 J) is her final kinetic energy. (d) Since K = mv2/2, her final speed is v= 2K/m = 2(1.06 x 103 J)/(72 kg) = 5.4 m/s 25. Givens: Figure 7-35, mC = 0.250 kg, mE = 900 kg, d1 = 2.40 m, d2 = 10.5 m, F is force exerted by cable, and (a) FN = 3.00 N and (b) WCable = 92.6 kJ. (a) The net force on the cheese is FN - mg = 3.00 N - (0.250 kg)(9.80 m/s2 ) = 0.55 N. So, the acceleration is a = F/m = (0.55 N)/(0.250 kg) = 2.2 m/s2 . The net force on the elevator is then F - Mg - FN = Ma = (900 kg)(2.2 m/s2 ) = 1980 N. i.e., F is acting upward, the weight of the elevator is downward and the normal force of the cheese on the floor of the elevator is downward. Solving for the force exerted by the cable, F = Mg + FN + ma = 8820 N + 3.00 N + 1980 N = 10803 N = 1.08 x 104 N, and the work done by the cable is Fd1 = (1.08 x 104 N)(2.40 m) = 2.59 x 104 J. (b) If W = 92.61 kJ and d2 = 10.5 m, then Fd2 = 9.261 x 104 J -> F = 8820 N. The net force on the cab is F - Mg = 0 -> a = 0. Afree-body diagram on the cheese would yield FN - mg = 0 -> FN = mg = 2.45 N. 28. Givens: Figure 7-11, F = 800 N, Wa = 4.0 J, xi = - 2.0 cm, v0 = v = 0. Find xf . We make use of Eqn. 7-25 and Eqn. 7-28 since the block is stationary before and after the displacement. The work done by the applied force can be written as Wa = - Ws = k(xf2 - xi2). The spring constant is k = (80 N)/(2.0 cm) = 4.0 x 103 N/m. With Wa = 4.0 J, and xi = - 2.0 cm, we have xi = (2Wa/k) + xi2 = (+/-) [2(4.0 J)/(4.0 x 103 N/m)] + (- 0.020 m)2 = (+/-)4.9 cm. 30. Givens: Figure 7-36 and xi = 12.0 cm. The work done by the spring force is given by Eqn. 7-25: Ws = k(xi2 - xf2)/2. Since Fx = - kx, the slope in Figure 7-35 corresponds to the spring constant k. Its value is given by k = 80 N/cm = 8.0 x 103 N/m. (a) When the block from from = xi + 8.0 cm to x = + 5.0 cm, we have Ws = (8.0 x 103 N/m)[(0.080 m)2 - (0.050 m)2]/2 = 15.6 J ~= 16 J. (b) Moving from xi = 8.0 cm to x = - 5.0 cm, we have Ws = (8.0 x 103 N/m)[(0.080 m)2 - (- 0.050 m)2 ]/2 = 15.6 J ~= 16 J. (c) Moving from xi = 8.0 cm to x = - 8.0 cm, we have Ws = (8.0 x 103 N/m)[(0.080 m)2 - (- 0.080 m)2 ]/2 = 0 J. (d) Moving from xi = 8.0 cm to x = - 10.0 cm, we have Ws = (8.0 x 103 N/m)[(0.080 m)2 - (- 0.10 m)2]/2 = - 14.4 J ~= - 14 J. 36. Givens: Figure 7-40, m = 10 kg. Find net work as brick moves from x = 0 cm to x = 8.0 cm. According to the graph, the acceleration a varies linearly with the coordinate x. We may write a = αx, where α is the slope of the graph. Numerically, α = (20 m/s2 )/(8.0 m) = 2.5 s-2. The force on the brick is in the positive x direction and, according to Newton’s second law, its magnitude is given F = ma = mαx. If xi is the final coordinate, the work done by the force is xf xf W = F dx = mα x dx = mαxf2 /2 = (10 kg)(2.5/s2)(8.0 m)2 = 8.0 x 102 J. 0 0 39. Givens: Figure 7-41 (quite similar, but not identical to Figure 5-59), m = 2.00 kg, x 0 = 0 and x = 9.0 m. (a) The approach is similar to that used in problem 5-67. In this case, the work done by Fa is equal to the area under the curve from x0 to x times the mass. For x = 4.0 m, this is (0.5)(6.0 m/s2 )(1.0 m) from 0 to 1 m plus (6.0 N)(3.0 m) from 1 to 4 m. The area under the curve is thus (3 + 18)(m/s)2 = 21 (m/s)2. Thus, the work done is (2.0 kg)(21 m2/s2) = 42 J. (b) From 4 to 6 m the work is 0 (since there is symmetry about x = 5 m). From 6 m to 7 m, the area is - 6 (m/s)2, and work is - 12 J, so the total work from 0 to 6 m is 30 J. (c) From 7 m to 8 m the work is also - 12 J and from 8 m to 9 m it’s - 6 J. The total work for the entire interval is thus (30 J - 12 J - 6 J) = 12 J. (d) Eqn. 7-10 (along with Eqn. 7-1) leads to the speed v = 6.5 m/s at x = 4.0 m. Returning to the original graph (where a was plotted) we note that (since it started from rest) it has received acceleration(s) (up up to this point) only in the + x direction and consequently must have a velocity vector pointing in the + x direction at x = 4.0 m. (e) Now, using the result of part (b) and Eqn. 7-10 (along with Eqn. 7-1) we find the speed is 5.5 m/s at x = 7.0 m. Although it experienced some deceleration during the 0 to 7 m interval, it velocity vector still points in the + x direction. (f) Finally, using the result from part (c) and Eqn. 7-10 (along with Eqn. 7-1) we find the speed v = 3.5 m at x = 9.0 m. It certainly has experienced a significant amount of deceleration during the 0 to 9 m interval; nonetheless, it velocity vector still points in the +x direction. The obvious way to confirm the above is to note that at no point has the total work become negative (up to this point). 49. Givens: m = 4.0 kg, di = [0.50 i + 0.75 j + 0.20 k]m at t = 0, df = [7.50 i + 12.0 j + 7.20 k]m at t = 12 s, and the constant force F = [2.00 i + 4.00 j + 6.00 k]N. (a) ∆r = df - di = [(7.50 i + 12.0 j + 7.20 k) - (0.50 i + 0.75 j + 0.20 k)]m = (7.00 i + 11.25 j + 7.00 k) Using W = F • ∆r (which we can use, since F is constant), we find W = F • ∆r = [7.00(2) + 11.25(4) + 7.00(6)]J = 101 J or 1.0 x 102 J. (b) Using the defining equation for average power (Eqn. 7-42) gives P = W/t = (100 J)/(12 s) = 8.4 W.