Work done in time t on a body of mass m which | vedclass

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(D) The body starts from rest,so initial velocity $u = 0$. The acceleration $a$ is constant,given by $a = \frac{v}{t_1}$.The force acting on the body

Vedclass · Accepted Answer

$\frac{1}{2}m\frac{v^2}{t_1^2}t^2$

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(D) The body starts from rest,so initial velocity $u = 0$. The acceleration $a$ is constant,given by $a = \frac{v}{t_1}$.The force acting on the body is $F = ma = m\left(\frac{v}{t_1}ight)$.The distance $s$ covered in time $t$ is $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}\left(\frac{v}{t_1}ight)t^2$.Work done $W$ is given by $W = F \cdot s$.Substituting the values,$W = \left(m\frac{v}{t_1}ight) 	imes \left(\frac{1}{2}\frac{v}{t_1}t^2ight) = \frac{1}{2}m\frac{v^2}{t_1^2}t^2$.

Work done in time $t$ on a body of mass $m$ which is accelerated from rest to a speed $v$ in time $t_1$ as a function of time $t$ is given by

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