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Question

In the given figure a block of mass m is tied on a wedge by an ideal string as shown in the figure. String is parallel towards to the inclined surface of wedge. All the surfaces involved are smooth. Wedge is being moved towards wight with a time varying velocity v=t22(m/s) where t is in sec. At what time block will just break the contact with wedge ( use g=10m/s )
1043655_3e2801abe92b4a939119b88601caa83a.png

A
10 sec
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B
5 sec
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C
2 sec
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D
4 sec
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Solution

The correct option is A 10 sec

The block will lift once the pseudo force acting on the wedge becomes equal to the weight of the block.

As the wedge is moving with the velocity oft22m/s, its acceleration will become

ddt(t22)=tm/s2.

The component of weight along the surface is equal to mgcos45.

The pseudo force acting opposite to it will be mtsin45

Both the forces will be equal at the time when the block lifts off the wedge.

mtsin45=mgcos45

t=g

So at time t=10sec the block will lift from the wedge.


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