Question

A body falling freely under gravity, hasits P.E. at a particular instant nine times its K.E, If the velocity of the body becomes double, the ratio of its K.E and P.E. will be *

1/2

2/3

4/5

1/4​

Answer 1

Answer:

1/2

Explanation:

Let a body of mass m, and it is falling from a hight h. Let, after falling a distance d , it's potential energy is nine times it's kinetic energy .

from a hight h , falling a distance d , it's hight is (h-d) , so potential energy = mg(h-d) , and let it's velocity v , so v^2 =2 g d ……..(1) so kinetic energy = 0.5 m ×(2gd) =mgd

as per condition 9mgd=mg(h-d)

or, 9d=h-d , or, 10d=h ……..(2)

let from the rest after falling a hight d' , it's velocity = 2v

(2v)^2= 2g d' , or, 4v^2=2gd'

or, 4gd=2gd' , as no(1)

or, 2d=d'

now it is at a hight (h- 2d)

=(10d-2d) =8d , as no (2)

and velocity 2v

kinetic energy= 0.5 m (2v)^2

potential energy=m g ×8d

ratio of Ke: Pe

= 2mv^2 : 8mgd

=2m× 2gd : 8mgd , as no (1)

=4mgd:8mgd

=1:2

Ke : Pe = 1:2