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Question

In the adjoining figure, a wedge is fixed to an elevator moving upwards with an acceleration a. A block of mass m is placed over the wedge. Find the acceleration of the block with respect to wedge. Neglect friction.

245549_aa5b34c68003427ebd18a042dc8dd7e9.png

A
(g+a)sinθ down the wedge
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B
gsinθ down the wedge
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C
(ga)sinθ down the wedge
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D
(ag)sinθ up the wedge
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Solution

The correct option is A (g+a)sinθ down the wedge
Free Body Diagram of block with respect to wedge is shown in Figure.
The acceleration would had been gsinθ (down the plane ). If the lift were stationary or when only weight (i.e., mg) acts downwards.
Here, downward force is m(g+a)
Therefore acceleration of the block with respect to wedge will be (g+a)sinθ down the plane.

1428706_245549_ans_1f887d90630e4ce3882081008c2e91b5.PNG

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