An engine pumps up $100 \ kg$ of water through a height of $10 \ m$ in $5 \ s$. Given that the efficiency of the engine is $60\%$. If $g = 10 \ m \ s^{-2}$,the power of the engine is .............. $kW$.

$A$ block of mass $1\,kg$ is pushed up a surface inclined to the horizontal at an angle of $30^{\circ}$ by a force of $10\,N$ parallel to the inclined surface (see figure). The coefficient of friction between the block and the incline is $0.1$. If the block is pushed up by $10\,m$ along the incline,calculate:
$(a)$ Work done against gravity
$(b)$ Work done against the force of friction
$(c)$ Increase in potential energy
$(d)$ Increase in kinetic energy
$(e)$ Work done by the applied force

$A$ motor engine pumps $1800 \,L$ of water per minute from a well of depth $30 \,m$ and allows it to pass through a pipe of cross-sectional area $30 \,cm^2$. Then the power of the engine is (Acceleration due to gravity,$g=10 \,m/s^2$) (in $\,kW$)

$A$ ball of mass $m$ falls vertically from a height $h$ and collides with a block of equal mass $m$ moving horizontally with a velocity $v$ on a surface. The coefficient of kinetic friction between the block and the surface is $0.2$,while the coefficient of restitution $e$ between the ball and the block is $0.5$. There is no friction acting between the ball and the block. The velocity of the block decreases by:

It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass $1.00 \; g$ falling from a height $1.00 \; km$. It hits the ground with a speed of $50.0 \; m s^{-1}$. $(a)$ What is the work done by the gravitational force? $(b)$ What is the work done by the unknown resistive force?

Is it possible for the kinetic energy of an object not to increase when work is done on it? If so,when?