Question

a block can slide on a smooth inclined plane of inclination Φ(theta) kept on the floor of a lift. when the lift is descending with the a retardation a, the acceleration of the block relative to the incline is
1) (g+a)sinΦ
2) ( g-a)
3) gsinΦ
4) ( g-a)sin Φ
​

Answer 1

The acceleration of the block relative to the incline is   option 1) (g+a)sinΦ

Given,

acceleration due to gravity = g

retardation of lift = -a

angle of inclination = Ф

Since the lift is descending, the retardation will act against the acceleration due to gravity

=> Net acceleration vertically = g - (-a) = g+a

=> Net acceleration component along the inclination = (g+a)sinФ

Hence the acceleration of the block relative to the incline is  

option 1) (g+a)sinΦ

Answer 2

Answer:

is the correct answer is first option