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

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(D) The acceleration of the body is $a = \frac{v}{t_1}$.The force acting on the body is $F = m \cdot a = m \cdot \frac{v}{t_1}$.The distance covered b

Vedclass · Accepted Answer

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

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(D) The acceleration of the body is $a = \frac{v}{t_1}$.The force acting on the body is $F = m \cdot a = m \cdot \frac{v}{t_1}$.The distance covered by the body in time $t$ starting from rest is $s = \frac{1}{2} a t^2 = \frac{1}{2} \left( \frac{v}{t_1} \right) t^2$.Work done is defined as $W = F \cdot s$.Substituting the values,$W = \left( m \cdot \frac{v}{t_1} \right) \cdot \left( \frac{1}{2} \cdot \frac{v}{t_1} \cdot t^2 \right)$.Simplifying the expression,$W = \frac{1}{2} m \left( \frac{v}{t_1} \right)^2 t^2 = \frac{1}{2} m \frac{v^2}{t_1^2} t^2$.

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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