a 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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Aman gohel · Answer

Given,F=3ti^+5j^S=2t^2i^ - 5j^At time t=2s,F=6i^+5j^S=8i^-5j^Work=F.S =6×8(i^)+5×5(-j^) =48i^-25j^(Ans)

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a 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

a 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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a 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

a 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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