Download mP = 1.67 x 10-27 kg, a = 3.6 x 1015 m/s2, v0 = 2.4 x 107 m/s, ∆x

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.