Question

When a shot-put champion releases a $5.0\text{kg}$ shot, it is at $1.5\text{m}$ above the ground and traveling at $10\text{m/s}$. It reaches a maximum height of $5\text{m}$ above the ground. What are the total mechanical energy, kinetic energy and potential energy of the shot at its maximum height?

Answer 1

Mass of shot-put, $m=5\text{kg}$
initial height $=1.5\text{m}$
Potential energy at initial height $=mgh$
$=5\times 10\times 1.5=75\text{J}$
Speed at initial height, $v=10\text{m/s}$
Kinetic energy at this height $=\frac{1}{2}m{v}^2$ $=\frac{1}{2}\times 5\times {10}^2=250\text{J}$
Mechanical energy at height = KE + PE $=250+75=325\text{J}$

According to energy conservation, total energy of a body remains constant.
Hence, at maximum height mechanical energy will be same and is equal to $325\text{J}$.

Now, potential energy at maximum height ($H$)
PE $=mgH=5\times 10\times 5=250\text{J}$

Kinetic energy at maximum height = Total energy - Potential energy
$\Rightarrow \text{KE}=325-250=75\text{J}$

Hence, at maximum height:
Total energy $=325\text{J}$
Kinetic energy $=75\text{J}$
Potential energy $=250\text{J}$