$A$ car moving with a velocity of $10 \, m/s$ can be stopped by the application of a constant force $F$ in a distance of $20 \, m$. If the velocity of the car is $30 \, m/s$,it can be stopped by this same force in what distance (in $, m$)?

$A$ bus moving on a level road with a velocity $v$ can be stopped at a distance of $x$ by the application of a retarding force $F$. The load on the bus is increased by $25\%$ by boarding the passengers. Now, if the bus is moving with the same speed and if the same retarding force is applied, the distance travelled by the bus before it stops is

$A$ particle slides from the top of a smooth hemispherical surface of radius $R$ which is fixed on a horizontal surface. If it separates from the hemisphere at a height $h$ from the horizontal surface,then the speed of the particle is

$A$ ball at rest is dropped from a height of $12 \,m$. It loses $25 \%$ of its kinetic energy on striking the ground and bounces back to a height '$h$'. Then the value of '$h$' is (in $\,m$)

The work-energy theorem is valid in the presence of:

Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying a retarding force of the same magnitude. How would the distance moved by them before coming to rest compare?