Work and Energy Crisis - Solutions

A link to the textbook website problems for Chapter 6. The Physlet problems and practice problems are great!

Solutions to today's problems:

1. Using the generalized work energy principle:
Wncf = Wfriction = Kf - Ko + Uf - Uo = .5(2 kg)( 40^2 - 20^2)m^2/s^2 + 0 - (2 kg)(9.8 m/s^2)(80 m) = -368 J

Wfriction = Ffriction*delta_x*cos(180) so Ffriction = -4.6 N

2.
a) a = 40 N / 5 kg = 8 m/s^2
b) Area between F and delta_x = 180 J230 J
c) Wnet = Kf - Ko so Kf = .5*m*vf^2 = Wnet + Ko

So v_f = (2(Wnet + Ko)/m)^(1/2) = (2(180 J + 22.5 J)/5 kg)^(1/2) = 10.04 m/s

d) For it to stop at x = 9 meters, the final KE must be zero. This means that the net work from x = 7 to x = 9 must be equal to Kf - Ko = Ko, the kinetic energy at the speed we found in (c).

The net work is going to be the sum of the work of the net-force and the work of friction, given by Ffric*delta_x*(-1).

The work done by the force from x = 7 to x = 9 is .5*2m*40 N = 40 J.

40 J - Ffric*(2 m)*(-1) = -.5*(5 kg)*(10.04 m/s)^2

Ffric = 106 N

3.
a) With K + U = 25 J, and at a displacement of 4 meters, U = 16 J. Thus K = 9 J

b) K = 1/2*m*v^2 so v = (2*K/m)^(1/2) = 3 m/s

c) When the object has maximum speed, it has maximum KE. This means it must have minimum U, and U has a minimum value of 0 here.

Therefore K_max = 25 J = 1/2*m*v^2

v = 5 m/s

d) The maximum displacement will occur when KE = 0, which occurs when U = 25 J. This occurs at x = +/- 5 meters.

4.
a) W_spring = .5*k*x^2 = 0.2 J
b) Only the spring force does work, so energy is conserved.

Ko + Uo = Kf + Uf
0 + 1/2*k*x^2 = 1/2*m*v^2 + 0
v = 6.32 m/s

c) Wncf = Kf - Ko + Uf - Uo
Ffric*delta_x*(-1) = 1/2*m*v_f^2 - 0 + 0 - 1/2*k*x^2

Ffric = .06 N