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A body of400g slides on a rough horizontal surface. If the frictional force is 3.0, the magnitude of contact force is
The answer is 5 newtonThe contact force on the body =(N2 + f2 )1/2N=Normal Force on the body which is 0.4kg*10m/s2=4Newtonf=frictional force is 3 Newton (given)therefore Contact Force is (42 + 32)1/2 = 5 Newton
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