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
m_k = 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
F_f and from figure with the drag force of 45-lb.
S F_y^' = ma_y^'
N - 75 cos30^o = (75/32.2) 0
N = 64.95 lb

Principle of work and energy : Force components parallel to the incilned plane 75 sin30^o lb and
F_f = m_{k N} = 0.5 (64.95) = 32.48 lb do the work,
where as the force components perpendicular to the inclined plane 75 cos30^o lb and normal reaction N do not work since no displacement occurs in this direction.
Here 75 sin30^o lb force does positive work and
F_f = 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 U_1-2 = T₂
0 + 75 sin30^o(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