Question

A body weighing 64𝑙𝑏 slides down from rest at the top of a plane 18𝑓𝑡 long and inclined 30𝑜 above the horizontal. The coefficient of friction is 0.10. Find the velocity of the body as it reaches the bottom of the plane.

Answer 1

The work-energy theorem gives:

$$\frac{mv^2}{2}=W_1+W_2=-\mu mgl\cos\theta+mgl\sin\theta$$

Hence, the speed of a body

$$v=\sqrt{2gl(-\mu \cos\theta+\sin\theta)}\\ =\sqrt{2*32.2*18*(-0.10 \cos30^{\circ}+\sin30^{\circ})}\\ =22\:\rm ft/s$$