A block is placed on the top of a smooth inclined plane of inclination θ kept on the floor of a lift. When the lift is descending with a retardation a, the block is released. The acceleration of the block relative to the incline is
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a
g sinθ
b
a sinθ
c
(g-a) sinθ
d
(g+a) sinθ
answer is C.
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Detailed Solution
Since the lift is accelerating downward , vertically upward pseudo force ma will be acting on the block and hence resultant force acting on the block along the incline relative to the lift is mg-asin θ .Hence, the acceleration of the block relative to the incline is (g-a) sinθ.