Question

a stone is dropped from a height h its potential energy at a particular instant is nine times its kinetic energy when the velocity is double what will be the ratio of its new kinetic energy and new potential energy ​

Answer 1

answer : 2 : 3

explanation : a stone is dropped from a height h. at a particle instant, stone is located at x from the top.

and potential of stone at x = 9 × kinetic energy at x.

initial velocity of stone , u = 0

so, v² = 0 + 2(-g){-x}

v² = 2gx

now, mg(h - x) = 9 × 1/2 m(2gx)

or, h - x = 9x

or, x = h/10.....(1)

so, old kinetic energy = mgh/10

and old potential energy = 9mgh/10

a/c to question,

velocity is doubled.

so, kinetic energy will be 4 times of old kinetic energy.

i.e., new kinetic energy = 4 × mgh/10

and then, new potential energy = mgh - 4mgh/10 = 6mgh/10

ratio of new kinetic energy and new potential energy = {4mgh/10}/{6mgh/10}

= 2 : 3

Answer 2

Answer:

A stone is dropped from a height h its potential energy at a particular instant is nine times its kinetic energy when the velocity is double what will be the ratio of its new kinetic energy and new potential energy ​is 2:3

Explanation:

The Potential Energy (PE) at a height h is given as = m*g*h

Where,

m is mass

g is acceleration due to gravity

h is the height

Let the height descended be "x"

Therefore the PE lost would be m*g*(h-x)

This PE lost will be converted to Kinetic Energy (KE)

Therefore,

KE = m*g*x

It is given that at a particular instant PE is 9 times KE

Therefore,

PE = 9*KE

m*g*(h-x)=9*m*g*x

h-x=9x

10x=h

x=h/10

When we double the velocity then the KE increases by 4 times

Hence,

KE now is = 4*KE

                = 4*m*g*x= 4*m*g*h/10= 2/5*m*g*h

But we know Total energy (TE)= PE at the top of the height i.e. mgh

therefore

By energy conservation,

PE= TE-KE

PE=mgh-2/5mgh

PE=3/5mgh

Therefore the ratio of the new KE to PE energy is (2/5):(3/5) i.e. 2:3