Question

A body of mass 12.0kg is placed on an orange tree 20.0m above the ground.

1. Calculate it's potential energy with respect to the ground.

2. If the body is released from thos height,what will be it's kinetic energy just before it hits the ground.

Answer 1

Explanations & Calculations

1.

• Potential energy is given by $\small E_p = mgh$. And the $\small h$ here is the height to the object measured with respect to a reference point with potential energy zero. The ground is considered the level of zero potential energy so this case is straightforward.

$\qquad\qquad \begin{aligned} \small E_p&=\small mgh\\ &=\small 12.0\,kg\times9.8\,ms^{-2}\times20.0\,m\\ &=\small 2352.0\,J \end{aligned}$

2.

• Conservation of mechanical energy is possible here as we think there exists no air resistance and so.
• The potential energy it had is transformed fully into kinetic energy at the ground level as the ground is zero potential energy.

$\qquad\qquad \begin{aligned} \small E_p+E_k&=\small E_p+E_k\\ \small mgh+0&=\small 0+E_k\\ \small E_k&=\small mgh=2352.0\,J \end{aligned}$