Question 113: Work, Energy and Power & Circular Motion

Question

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 :

Accepted Answer

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

Answer

$$-$$0.68 mgh

Answer

mgh

Answer

1.64 mgh

Answer

0.64 mgh

Work, Energy and Power & Circular Motion

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 :

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