Answer to Question #239480 in Physics for rshiii

Question

4. A block of mass 3.0 kg slides with uniform velocity down a plane inclined 25o with the horizontal. If the angle of inclination is increased to 40o , what will be the acceleration of the block? 

Answer 1

Given:

$\theta_1=25^{\circ},\quad \theta_2=40^{\circ}$

The Newton's second law gives

$$a=g(\sin\theta-\mu\cos\theta)$$

The coefficient of kinetic friction is given by

$$\mu=\tan\theta_1$$

Hence, the acceleration of the body

$$a=g(\sin\theta_2-\tan\theta_1\cos\theta_2)$$

$$a=9.8\:{\rm m/s^2}(\sin40^{\circ}-\tan25^{\circ}\cos40^{\circ})=2.8\:\rm m/s^2$$