Question

(1). A stone of l kg dropped from a height 100m reaches
the ground with a speed 40 m/s. Calculate the work done by the viscus drag of air on the stone.​

Answer 1

Given:

Stone of 1 kg is dropped from a height of 100 m and reaches the ground with a speed of 40m/s.

To find:

Work done by viscous drag of Air on the stone.

Calculation:

In this type of questions, it is best to apply work energy theorem which states that the net work done by all the forces will be equal to change in kinetic energy of the body.

$\sf{W_{air drag} + W_{gravity} = \Delta KE }$

$\sf{ = > W_{air drag} + mgh= \dfrac{1}{2} m {v}^{2} }$

$\sf{ = > W_{air drag} + (1 \times 10 \times 100)= \dfrac{1}{2} \times 1 \times {40}^{2} }$

$\sf{ = > W_{air drag} + 1000= 800}$

$\sf{ = > W_{air drag} = - 200}$

$\sf{ = > W_{air drag} = - 200 \: Joule}$

Negative Work refers to resistance to movement provided by air drag.

Final answer is:

$\boxed{ \red{ \large{\rm{ W_{air drag} = - 200 \: Joule}}}}$

Answer 2

A stone of mass 1kg dropped from a height 100m reaches the ground with a speed of 40 m/s.

To find : The work done by the viscus drag of air on the stone.

solution : from conservation of work - energy theorem,

at a height 100m, stone has potential energy. after dropping, potential energy is converted into kinetic energy and work done by the air drag on the stone.

i.e., potential energy = kinetic energy + work done by the air drag

⇒mgh = 1/2 mv² + work done by the air drag

here m = 1 kg, h = 100 m and v = 40 m/s²

⇒1 kg × 10 m/s² × 100 m = 1/2 × 1 kg × (40 m/s)² + work done by the air drag

⇒work done by the air drag = 1000 J - 800 J = 200 J

Therefore the work done by the air drag on the stone is 200 J.