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Question

Physics Question on work, energy and power

A plate of mass mm, length bb and breadth aa is initially lying on a horizontal floor with length parallel to the floor and breadth perpendicular to the floor. The work done to exect it on its breadth is

A

mg[b2]mg\Bigg[\frac{b}{2}\Bigg]

B

mg[a+b2]mg\Bigg[a+\frac{b}{2}\Bigg]

C

mg[ba2]mg\Bigg[\frac{b-a}{2}\Bigg]

D

mg[b+a2]mg\Bigg[\frac{b+a}{2}\Bigg]

Answer

mg[ba2]mg\Bigg[\frac{b-a}{2}\Bigg]

Explanation

Solution

Initial height of centre of gravity=a2=\frac{a}{2} Fmal heig t o centre o gravity =b2=\frac{b}{2} Work done=mg[b2a2]=mg\Bigg[\frac{b}{2}-\frac{a}{2}\Bigg] =mg[ba2]\, \, \, \, \, \, \, \, \, =mg\Bigg[\frac{b-a}{2}\Bigg]