Question

q4 a 2kg ball is dropped from a height of 10m
a. find the potential energy at the top.
b. find its kinetic energy just before hitting the ground

Explanation:

Step1: Calculate potential energy

Use formula \(PE = mgh\). Here \(m = 2\space kg\), \(g= 9.8\space m/s^{2}\), \(h = 10\space m\).
\(PE=2\times9.8\times10\)

Step2: Use conservation of energy

At the top, initial kinetic energy \(KE_{i}=0\) (dropped, so initial velocity \(v = 0\)). Just before hitting the ground, height \(h = 0\), so potential energy \(PE_{f}=0\). By conservation of mechanical energy \(E_{i}=E_{f}\), \(PE_{i}+KE_{i}=PE_{f}+KE_{f}\). Since \(PE_{i}=196\space J\), \(KE_{i} = 0\), \(PE_{f}=0\), then \(KE_{f}=PE_{i}\)

Answer:

a. The potential energy at the top is \(196\space J\)
b. The kinetic energy just before hitting the ground is \(196\space J\)