$A$ body falls freely under gravity. Its speed is $v$ when it has lost an amount $U$ of the gravitational potential energy. Then its mass is

Derive the work-energy theorem for a variable force exerted on a body in one dimension.

$A$ 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 $\,J$)

The energy required to accelerate a car from $10 \ m/s$ to $20 \ m/s$ is how many times the energy required to accelerate the same car from rest to $10 \ m/s$?

$A$ body of mass $0.25 \ kg$ travels along a straight line from $x=0$ to $x=2 \ m$ with a speed $v=k x^{3/2}$ where $k=2$ $SI$ units. The work done by the net force during this displacement is: (in $J$)

$A$ force $\overrightarrow{F} = 4 \hat{i} - 15 \hat{j} \text{ N}$ acts on a body, resulting in a displacement of $\overrightarrow{S} = 6 \hat{i} \text{ m}$. If the body had a kinetic energy of $7 \text{ J}$ at the beginning of the displacement, what is the kinetic energy at the end of the displacement (in $\text{ J}$)?