A cricket ball of mass $0.15\, kg$ is thrown vertically up by a bowling machine so that it rises to a maximum height of $20 \;m$ after leaving the machine. If the part pushing the ball applies a constant force $F$ on the ball and moves horizontally a distance of $0.2\, m$ while launching the ball, the value of $F($ in $N)$ is
$\left(g=10\, m s^{-2}\right)$
$200$
$150$
$275$
$325$
A block of mass $m$ is sliding down an inclined plane with constant speed.At a certain instant $t_0$, its height above the ground is $h$. The coefficient of kinetic friction between the block and the plane is $\mu$. If the block reaches the ground at a later instant $t_g$, then the energy dissipated by friction in the time interval $\left(t_g-t_0\right)$ is
A body of mass $m= 10^{-2}$ $kg$ is moving in a medium and experiences a frictional force $F= -kv^2$. Its initial 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
Derive the work energy theorem for a variable force exerted on a body in one dimension.
A force acts on a $2\,kg$ object so that its position is given as a function of time as $x= 3t^2 + 5.$ What is the work done by this force in first $5\,seconds$ ? ................ $\mathrm{J}$
Consider an elliptically shaped rail $P Q$ in the vertical plane with $O P=3 \ m$ and $OQ =4 \ m$. A block of mass $1 \ kg$ is pulled along the rail from $P$ to $Q$ with a force of $18 \ N$, Which is always parallel to line $PQ$ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches $Q$ is $(n \times 10)$ joules. The value of $n$ is (take acceleration due to gravity $=10 \ ms ^{-2}$ )