Question

A body dropped from h height reaches the ground with speed 1.2 root 2gh find work done by air

Answer 1

Potential energy at highest point = mgh

kinetic energy when ball reaches to ground

= m = mgh

1

2 v

2 3

4

Change in energy = work done by the air = mgh − mgh

3

4

=

mgh

Answer 2

The work done by air friction is - 0.28 mg H.

Explanation :

Given :

A body of mass 'm' dropped from a height 'H' and reaches the ground with a speed of $\bf 1.2\;\sqrt{gh}$

u - 0 m/s.

v - $\bf 1.2\;\sqrt{gh}$

To Find :

The work done by air friction.

Solution :

We know the Work-Energy Theorem :

So,

Total work done = Total change in Kinetic Energy.

So,

$\implies \sf K.E. = \dfrac{1}{2}\;mv^2- \dfrac{1}{2}\;mu^2\\ \\ \\\implies \sf W = \dfrac{1}{2} m(1.2\sqrt{gH}^2 - \dfrac{1}{2} m(0)^2\\ \\ \\\implies \sf W = \dfrac{1}{2} m(1.44gH)\\ \\ \\\implies \sf W = 0.72mgH$

&

$\textbf{Work done by gravity = mgH \bigg[(-m)(-gH)\bigg]}$

Now,

Total work done = Work done by gravity + Work done by air friction.

Hence,

Work done by air friction = Total work done - Work done by gravity.

So,

$\implies \sf 0.72mgH - mgH\\ \\ \\ \implies \sf mgH(0.72 - 1)\\ \\ \\ \implies \sf mgH(-0.28)\\ \\ \\ \implies \sf -0.28mgH$

Therefore, the work done by air friction is - 0.28 mg H.