Let a crate of 75 lb be dragged untill half the plane as shown and be released from that point. Determine its speed after it slides down the plane. The coefficient of kineic friction between the crate and plane is
mk = 0.5.

1) 11.4 ft/s
2) 19.0 ft/s
3) 23.5 ft/s
4) 30.3 ft/s

Equation of motion: Since the crate slides, the friction force developed between the crate and its surface is
Ff and from figure with the drag force of 45-lb.
S Fy' = may'
N - 75 cos30o = (75/32.2) 0
N = 64.95 lb

Principle of work and energy : Force components parallel to the incilned plane 75 sin30o lb and
Ff = mk N = 0.5 (64.95) = 32.48 lb do the work,
where as the force components perpendicular to the inclined plane 75 cos30o lb and normal reaction N do not work since no displacement occurs in this direction.
Here 75 sin30o lb force does positive work and
Ff = 32.48 lb does the negative work. Since the crate is originally at rest T1 = 0.
Applying the conservation of energy principle we have:
T1 + S U1-2 = T2
0 + 75 sin30o(30) - 32.48(30) = 1/2(75/32.2) v2
v2 = 129.71 (ft/s)2
v = 11.38 ft/s

So the speed of the crate after it slides down the plane is v = 11.4 ft/s