force F is equal to 3ti^ + 5j^ in acts on a body due to which its position varies as S is equal to 2t^2i^ - 5j^. Work done by this force in first 2 seconds is

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Arun · Answer

Dear Vivek Work done = F. dS = (3t i + 5j).(2 t² i - 5j) = 6t³ -25 At t= 2 sec Work done = 6 *2³ -25 = 48- 25 = 23 Joules Regards Arun (askIITians forum expert)

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force F is equal to 3ti^ + 5j^ in acts on a body due to which its position varies as S is equal to 2t^2i^ - 5j^. Work done by this force in first 2 seconds is

force F is equal to 3ti^ + 5j^ in acts on a body due to which its position varies as S is equal to 2t^2i^ - 5j^. Work done by this force in first 2 seconds is

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force F is equal to 3ti^ + 5j^ in acts on a body due to which its position varies as S is equal to 2t^2i^ - 5j^. Work done by this force in first 2 seconds is

force F is equal to 3ti^ + 5j^ in acts on a body due to which its position varies as S is equal to 2t^2i^ - 5j^. Work done by this force in first 2 seconds is

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