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Q.

A block of ‘m’ is placed at the bottom of a mass less smooth wedge which it placed on a horizontal force.  When it pushes the wedges with a force block moves.  Find the work done, when the block has a speed ‘v’ and is reaches the top of wedge.

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a

mgy

b

–mgy

c

12mv2

d

12mv2+mgy

answer is D.

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Detailed Solution

Given mass of block = mAccording  to work energy theorem Wext+Wgravity=ΔKE                (∵Wgravity=mgy)Wext−mgy=ΔKEWext=12mv2+mgy
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