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
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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