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
$\frac{1}{2}m\,\frac{v}{{{t_1}}}{t^2}$
$m\,\frac{{{v}}}{{{t_1}}}{t^2}$
$\frac{1}{2}{\left( {\,\frac{{mv}}{{{t_1}}}} \right)^2}{t^2}$
$\frac{1}{2}m\,\frac{{{v^2}}}{{{t^2}_1}}{t^2}$
A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time $t^{\prime}$ is proportional to :
The total work done on a particle is equal to the change in its kinetic energy. This is applicable
A body of mass $m= 10^{-2} \;kg$ is moving in a medium and experiences a frictional force $F= -kv^2$ Its intial speed is $v_0= 10$ $ms^{-1}$ If, after $10\ s$, its energy is $\frac{1}{8}$ $mv_0^2$ the value of $k$ will be
A bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be
A body of mass $m$ is moving in a circle of radius $r$ with a constant speed $u$. The force on the body is $mv^2/r$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle?