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A block of mass 2 kg is placed on a rough horizontal surface, coefficient of friction between the block and the surface is 0.4. A horizontal force of 10N is applied on the block. Then rate of work done by the force at t = 1 sec is (The 10 N force is applied at t = 0).

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Detailed Solution

a=10−0.4 × 2 ×10 2 m/s2=1m/s2After 1 sec, v = u + at = 0 + 1 x 1 m/s2 = 1 m/s∴ Power P = F.ν = 10 x 1 Watt = 10 Watt

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A block of mass 2 kg is placed on a rough horizontal surface, coefficient of friction between the block and the surface is 0.4. A horizontal force of 10N is applied on the block. Then rate of work done by the force at t = 1 sec is (The 10 N force is applied at t = 0).