A block is placed on top of a smooth inclined plane of inclination θ kept on the floor of a lift. When the lift descends with a retardation a, the relative acceleration of the block, parallel to the surface of the slope is ......... .
(A) g sinθ
(B) asinθ
(C) (g - a)sinθ
(D) (g + a)sinθ
Answers
Answered by
1
Answer:
Option (D) (g + a)sinθ
Explanation:
When the lift is descending, the retardation of the lift -a will be downwards
The acceleration due to gravity g will also be downwards
In the frame of reference of the lift, the net acceleration acting on the block downwards will be g - (- a) = g + a
The component of this acceleration parallel to the plane = (g + a)sinθ
Hope this is helpful.
Similar questions