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the displacement x of a body of mass 1 kg on smooth horizontal surface as a function of time T is given by X equal tq by 3 where X is in metre and T is in second find the work done by the external agent for the first one second |
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Answer» ^2 DX / dt = v = 4T d2x / DT2 = a = 4 Force acting on a body = m*a = 1 * 4 = 4 Newton W = \int F . dx w = \int 4 * 4t dt = 16 t^2 / 2 = 8 t^2 now put t = 1hence w = 8 JOULES |
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