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Question

A block of mass m is released from top of a smooth fixed inclined plane of inclination $θ$ . Find out work done by normal force.

0
$m g h$
$2 m g h$
$4 m g h$

A

0

B

2 m g h

C

4 m g h

D

m g h

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Solution

Verified by Toppr

The normal force will act in the direction normal to the inclined plane whereas the displacement of the block is along the inclined plane.

Angle made b/w normal force and displacement of the block $= ϕ = 90^{o}$

Now, Work done $= W = F s cos ϕ = N \times s \times cos 90^{o} = 0$

Hence, correct answer is option $A$

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