A body of mass m accelerates uniformly from r | vedclass

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(A) The body starts from rest,so initial velocity $u = 0$. Since the acceleration $a$ is uniform,$a = \frac{v_0 - u}{t_0} = \frac{v_0}{t_0}$. The forc

Vedclass · Accepted Answer

$\frac{1}{2} m v_0^2 \left( \frac{t^2}{t_0^2} \right)$

Vedclass · Answer

(A) The body starts from rest,so initial velocity $u = 0$. Since the acceleration $a$ is uniform,$a = \frac{v_0 - u}{t_0} = \frac{v_0}{t_0}$. The force acting on the body is $F = ma = m \left( \frac{v_0}{t_0} \right)$. The distance $S$ covered in time $t$ is given by $S = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \left( \frac{v_0}{t_0} \right) t^2$. The work done $W$ is equal to the product of force and displacement: $W = F \cdot S = \left( m \frac{v_0}{t_0} \right) \left( \frac{1}{2} \frac{v_0}{t_0} t^2 \right)$. Simplifying this,we get $W = \frac{1}{2} m v_0^2 \left( \frac{t^2}{t_0^2} \right)$.

$A$ body of mass $m$ accelerates uniformly from rest to a speed $v_0$ in time $t_0$. The work done on the body till any time $t$ is

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