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