$A$ body is projected vertically upwards with a velocity of $20 \,m \,s^{-1}$. If the potential energy of the body at a height of $5 \,m$ from the ground is $100 \,J$, then the kinetic energy of the body at a height of $10 \,m$ from the ground is (Acceleration due to gravity $= 10 \,m \,s^{-2}$) (in $\,J$)

$A$ 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

$A$ spherical ball of mass $20 \, kg$ is stationary at the top of a hill of height $100 \, m$. It slides down a smooth surface to the ground,then climbs up another hill of height $30 \, m$ and finally slides down to a horizontal base at a height of $20 \, m$ above the ground. The velocity attained by the ball is ............... $m/s$.

$A$ body is released from a height of $30 \ m$ vertically downwards. The speed of the body at which potential energy is twice that of kinetic energy is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

$A$ simple pendulum is released from $A$ as shown in the figure. If $m$ and $l$ represent the mass of the bob and the length of the pendulum respectively,the gain in kinetic energy at $B$ is

$A$ body of mass $0.5\, kg$ travels on a straight-line path with velocity $v = (3x^2 + 4)\, m/s$. The net work done by the force during its displacement from $x = 0$ to $x = 2\, m$ is $.......J$.