Answer to Question #274927 in Physics for Dhea

Question

A 60N block is resting at the top of a 30 degree slope of height 2m. It is attached to a thin walled cyliindrical pulley of mass 5 kg and radius 30 cm by a massless string that unwinds as the block slides downhill. If the acceleration of the block is 2.5 rad/second square, what is the value of the coefficient of kinetic friction? What is the velocity at the bottom of the slope?

Answer 1

Find the linear acceleration of the block:

$$a=\alpha r=2.5·0.3=0.75\text{ m/s}^2.$$

The coefficient of kinetic friction:

$$\mu=\tan\theta-\frac{a}{g\cos\theta}=0.49.$$

Velocity at the bottom:

$$v=\sqrt{2ah/\sin\theta}=2.4\text{ m/s}.$$