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Q.

A grinding machine whose wheel has a radius of 1π is rotating at 2.5 rev/sec. A tool to be sharped is held against the wheel with a force of 40N. If the coefficient of friction between the tool and wheel is 0.2, power (in W) required is

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answer is 40.

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Detailed Solution

The normal force is given by F=40N

So, the frictional force is given by f=0.2×F=8N

Power=Frictional force x velocity of the wheel.

P=frω =8×1π×2.5×2π=40

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