The decrease in potential energy between top position A and bottom position B is- - m g r - - m g r - 2 m g r - 1-63 This must be equal to the increase in kinetic energy- when particle move from A to B i-e- 1 - 2 m v 2 2 - 1 - 2 m v 1 2

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Standard XII

Physics

Introduction

The decrease ...

Question

The decrease in potential energy between top position A and bottom position B is,
$= m g r - (- m g r) = 2 m g r \dots (1.63)$
This must be equal to the increase in kinetic energy, when particle move from A to B
i.e. $\frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$

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Solution

From the equation,
$m g h =$ potential energy
$m g r - m g r = 0$
But total $m r g - (- m g r)$ $= 2 m g r$
and kinetic energy $\frac{1}{2} m (V_{2}^{2} - V_{1}^{2})$

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The decrease in potential energy between top position A and bottom position B is, =mgr−(−mgr)=2mgr…(1.63) This must be equal to the increase in kinetic energy, when particle move from A to B i.e. 12mv22−12mv21

From the equation,mgh= potential energymgr−mgr=0But total mrg−(−mgr) =2mgrand kinetic energy 12m(V22−V21)

The decrease in potential energy between top position A and bottom position B is,
$= m g r - (- m g r) = 2 m g r \dots (1.63)$
This must be equal to the increase in kinetic energy, when particle move from A to B
i.e. $\frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$

From the equation,
$m g h =$ potential energy
$m g r - m g r = 0$
But total $m r g - (- m g r)$ $= 2 m g r$
and kinetic energy $\frac{1}{2} m (V_{2}^{2} - V_{1}^{2})$