Question

a metal ball weighing 5 N is thrown vertically upward with a velocity of 39.2m/s. calculate it's potential energy and the maximum height reached using conservation of energy principle

Answer 1

Given:

A metal ball weighing 5 N is thrown vertically upward with a velocity of 39.2m/s.

To find:

• Potential energy?

• Max height reached ?

Calculation:

As per the question, we will use CONSERVATION OF MECHANICAL ENERGY principle :

$\sf \therefore \: KE1 + PE1 = KE2 + PE2$

$\sf\implies \: \dfrac{1}{2} m {u}^{2} + 0 = PE + 0$

$\sf\implies \: \dfrac{1}{2} m {u}^{2} = PE$

$\sf\implies \: PE = \dfrac{1}{2} \times \dfrac{weight}{g} \times {u}^{2}$

$\sf\implies \: PE = \dfrac{1}{2} \times \dfrac{5}{9.8} \times {(39.2)}^{2}$

$\boxed{ \sf\implies \: PE = 392 \: joule}$

$\sf\implies \: mgh = 392 \: joule$

$\sf\implies \: 5 \times h = 392 \: joule$

$\boxed{ \sf\implies \: h = 78.4 \: m}$

Hope It Helps