Question

A block of wood resting on an inclined plane of angle 30° starts moving down. If coefficient
friction is 0.2, its velocity (in m/s) after 5 sec is (g=10m/s^):-
(1) 12.75
(2) 16.34
(3)18.25
(4) 20​

Answer 1

Answer:

what do we have to find in this question?And what is the velocity after 5 sec

Answer 2

Answer:

A block of mass 10kg is moving on inclined plane with constant velocity . means body doesn't move with acceleration. hence body is in inertial frame. so, we have to treat as equilibrium condition.

Let angle of inclination of plane is \theta

so, component of weight along plane = mgsin\theta

component of weight perpendicular to plane = mgcos\theta

at equilibrium,

components of weight along plane = friction

mgsin\theta = \muN

where \mu is coefficient of friction and N is normal reaction e.g., N = mgcos\theta

so, mgsin\theta=\mu mgcos\theta

\mu=tan\theta

hence, coefficient of friction = tan