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Express kinetic energy by calcalus method

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Express kinetic energy by calcalus method
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Shabya Shaikh 1 year, 3 months ago

Suppose, body is initially at rest and force 'F' is applied on the body to displace it through 'ds' along it's own direction. Then small workdene, dw = f.ds dw = fds Acc. to Newton's 2nd law of motion f = ma Therefore, dw = ma ds dw = m dv/dt ds dw = m ds/dt dv dw = mvdv Integrating both sides In order to increase it's velocity from 0-v workdone is given by -- w = m sign of integration upper limit v and lower limit 0 vdv w = m[v square/2] upper limit v and lower limit 0 w = m[v square/2-0] w = 1/2mv square workdone is stored in form of energy Therefore, K.E. = 1/2mv square
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