A 45-lb force is applied to drag a crate 30 ft up the rough inclined plane at constant velocity. If the force becomes zero when the crate reaches the top, and the crate is then released from rest, determine the crate's velocity when it slides back down the plane and reaches the bottom if the mass of the crate is 100-lb.

1) 27.2 ft/s
2) 39.0 ft/s
3) 43.5 ft/s
4) 35.3 ft/s

Equation of motion: Since the crate slides, the friction force developed between the crate and its surface is
F_f and from fig (a) with the drag force of 45 lb.
S F_x^' = ma_x^'
45 - 100 sin24.62^o - F_f = (100/32.2) 0
F_f = 45 - 41.66 = 3.34 lb

Principle of work and energy : From the figure (b) with no drag force 100 sin24.62^o and F_f = 3.34 lb do the work whereas 100 cos24.62^o and normal reaction N do not work since no displacement occurs in this direction.
Here 100 sin24.62^o does positive work and
F_f = 3.34 lb does the negative work. Applying the conservation of energy principle we have:
T₁ + S U_1-2 = T₂
0 + 100 sin24.62^o(30) - 3.34(30) = 1/2(100/32.2) v²
1249.8 - 100.2 = 1/2(100/32.2) v²
v² = 740.33 (ft/s)²
v = 27.21 ft/s

So the velocity of the crate when it slides back the plane and reaches ponit A is v = 27.2 ft/s