Question

A box of mass 2.00kg is released from rest at the top of a plane inclined 15.3º above horizontal. The coefficient of kinetic friction is 0.22. What will be the speed of the mass after sliding 4.2 m along the plane? Round to the nearest hundredth.

Answer 1

1) Determine the acceleration:

$$a=g(\sin\theta+\mu\cos\theta).$$

2) Find the speed:

$$v=\sqrt{2ad}=\sqrt{2dg(\sin\theta+\mu\cos\theta)},\\ v=6.26\text{ m/s}.$$