Answer to Question #244438 in Physics for Kristine

Question

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:

$\theta_1=25^{\circ}$

$\theta_2=40^{\circ}$

The friction cooficient

$$\mu=\tan\theta_1=\tan 25^{\circ}=0.466$$

The acceleration of the object on the inclined plane

$$a=g(\sin\theta_2-\mu\cos\theta_2)$$

Hence,

$$a=9.81*(\sin40^{\circ}-0.466*\cos40^{\circ})=2.80\:\rm m/s^2$$