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A ball of mass m is dropped from a hight H At height H/3, the ratio of its potential energy to kinetic energy is equal to


Adarsh gautam 19th Feb, 2019
Answer (1)
Pulkit 20th Feb, 2019

Hello Adarsh,

According to the law of conservation of energy take the the ground as zero. Take point A as the point at a height H from the ground and point B at a height of h/3.

Now, as the mass m is dropped to h/3 i.e. to point B,  the height from the ground at that point will be H-h/3= 2h/3 .

At this point the potential energy will be mg(2h/3) = 2mgh/3

And the kinetic energy will be = 1/2mv^2

We have to find v at point B

At point A, u= 0 as the mass is just dropped

and a = g , s= h/3

So, by v^2 - u^2 = 2as

v^2 - 0 = 2×g×h/3

v^2 = 2gh/3

Now put the value of v^2 in the above equation i.e. 1/2 mv^2 we will get

Kinetic energy= 1/2m (2gh/3)

= mgh/3

Potential energy= 2mgh/3

Now the ratio of potential energy to kinetic energy will be

=P.E. / K.E.

= 2mgh/3 / mgh/3

= 2


Hope this will help you. Feel free to ask any query.

All the best!!


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