Average Potential Energy

Classical Mechanics Level pending

A ball of mass m m is thrown upward with some velocity. It reaches a maximum height of H H till the time T 2 \frac{T}{2} . If I choose a random moment of time between t = 0 t=0 to t = T 2 t=\frac{T}{2} what will be its expected gravitational potential energy? If your answer comes in the form of α m g H \alpha mgH Type your answer as α = ? \alpha=?


The answer is 0.667.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

The expectation value of the potential energy is the average value weighted over time from 0 0 to T 2 \dfrac{T}{2} :

<P. E.> T = 2 T ∫ 0 T 2 m g h d t = 2 m g T ∫ 0 T 2 ( u t − 1 2 g t 2 ) d t = 2 m g T ( u 2 × T 2 4 − g 6 × T 3 8 ) = m g ( u 2 2 g − u 2 6 g ) = m g ( H − H 3 ) = 2 3 m g H _T=\dfrac{2}{T}\displaystyle \int_0^{\frac{T}{2}} mghdt =\dfrac{2mg}{T}\displaystyle \int_0^{\frac{T}{2}} (ut-\frac{1}{2}gt^2)dt=\dfrac{2mg}{T}\left (\frac{u}{2}\times \frac{T^2}{4}-\frac{g}{6}\times \frac{T^3}{8}\right ) =mg\left (\frac{u^2}{2g}-\frac{u^2}{6g}\right ) =mg(H-\frac{H}{3})=\dfrac{2}{3}mgH . Hence, α = 2 3 ≈ 0.667 α=\dfrac{2}{3}\approx \boxed {0.667} .

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@Alak Bhattacharya Nice solution. I solved this through this method only. Is there any more method to solve this??

A Former Brilliant Member - 1 year, 1 month ago

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