A body of mass 'm' dropped from a height 'h' reaches the ground with a speed of 0.8$\sqrt {gh}$. The value of workdone by the air-friction is :

Given, the mass of the body = m<br/><br/>The height from which the body dropped = h<br/><br/>The speed of the body when reached the ground, ${v_f} = 0.8\sqrt {gh}$<br/><br/>Initial velocity of the body, v = 0 m/s<br/><br/>Using the work-energy theorem,<br/><br/>Work done by gravity + Work done by air-friction = Final kinetic energy $-$ Initial kinetic energy.<br/><br/>${W_{mg}} + {W_{air - friction}} = {1 \over 2}mv_f^2 - {1 \over 2}mv_i^2$<br/><br/>Here, work done by gravity = mgh<br/><br/>$$ \Rightarrow mgh + {W_{air - friction}} = {1 \over 2}m{(0.8\sqrt {gh} )^2} - {1 \over 2}m{(0)^2}$$<br/><br/>$\Rightarrow {W_{air - friction}} = {{0.64mgh} \over 2} - mgh$<br/><br/>$\Rightarrow 0.32mgh - mgh = - 0.68mgh$<br/><br/>The value of the work done by the air friction is $-$ 0.68 mgh.