Find the at which the kinetic energy is 1/4 the potential energy when a mass is dropped from a height of 20 metres. Also find the velocity at this point.

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Solution

Solution:

Given values:

Height, h = 20 m;

Let a body of mass 'm' be dropped from a height 'h' above the ground.
The potential energy of the body = mgh;
Take, Initial velocity of the body (u) = 0;

$K i n e t i c E n e r g y, K E = \frac{1}{2} m v^{2} v i s t h e v e l o c i t y w i t h w h i c h t h e b o d y c o v e r s a d i s \tan c e h; N o w, m a k e u s e o f t h e e q u a t i o n o f m o t i o n t o c a l c u l a t e v e l o c i t y, v; v^{2} - u^{2} = 2 g h v^{2} - 0 = 2 g h v^{2} = 2 g h K i n e t i c E n e r g y = \frac{1}{2} m v^{2} - - - - - - - (1) P u t v v a l u e i n e q (1) T h e n, K i n e t i c E n e r g y = \frac{1}{2} m x 2 g h K i n e t i c E n e r g y = m g h$

This shows that the kinetic energy of the body when it reaches the ground after travelling a height 'h' is equal to the potential energy of the body at height 'h'.

Now,
Calculate the point where the K.E. = 1/2 P.E. if any object fall from the height of 20 meter.

Kinetic Energy = Potential Energy = mgh

Then, if any object with mass 'm' fall from 20 meter, the point where Potential Energy is 4th time lessor then Kinetic Energy is equal to:

P.E = mgh = m x g x 20 or it divided by 4 then we get the point where Potential Energy 4th time lessor then KE is 5 Meter

Now,

The given values:

Height = 5 m;
g = 9.8 m/s²
m = constant;

Then,
$\frac{1}{2} m v^{2} = m g h v^{2} = 2 g h v = \sqrt{2 g h} v = \sqrt{2 x 9.8 x 5} v = \sqrt{98} v = 9.89 m / s;$

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