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

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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. $\frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$

Open in App

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 $\frac{1}{2} m (V_{2}^{2} - V_{1}^{2})$

Suggest Corrections

Similar questions

Q. The kinetic energy needed to project a body of mass m from the earth' surface to infinity is
(a)

\frac{1}{4} m g R

(b)

\frac{1}{2} m g R

(c)

m g R

(d)

2 m g R

Q. The change in Gravitational potential energy when a body of mass

m

is raised to a height

n R

from the Earth’s surface is (

R

= radius of the earth,

g

= acceleration due to gravity on the surface of Earth)

A carpet of mass $M$ is rolled along its length in the form of a cylinder of radius $R$ and kept on a rough floor. The decrease in potential energy, if the carpet is unrolled without sliding to a radius $\frac{R}{2}$ is equal to

Q. At two positions kinetic energy and potential energy of a particle are

K_{1} = 10 J : U_{1} = - 20 J, K_{2} = 20 J, U_{2} = - 10 J

In moving from 1 to 2,

Q. Find the work done to take a particle of mass

m

from surface of the earth to a height equal to

2 R

Join BYJU'S Learning Program

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 $\frac{1}{2} m (V_{2}^{2} - V_{1}^{2})$