Question

A 5 kg ball when falls through a height of 20 metre acqures a speed of 10 metre per second find the work done by the air resistance.​

Answer 1

answer:

from the work energy theorem work done by weight+work done by air = delta KE

W = delta KE -W (weight)

=1/2mv^2 -mgh

=1/2 × 5 × (10)^2 - 5 × 10 × 20

=-750J

HAPPY TO HELP

Answer 2

Answer;

-750J

Given

Speed of the ball (v) = 10m/s

Mass of the body (m) = 5

Height (h) = 20m

According to Question

The ball is starting falling from Position (P1), where it's speed is 0

Then

Kinetic energy will also be 0.

$K_{1} = 0J...................(i)$

During the downward motion of the ball, constant gravational force mg acts downward

And

A resistance R of unknown magnitude will act upward as we can see in the free body diagram.

Now

When the ball will reach at position 20m below the position (P1) with speed

v = 10m/s,

Then ,

The the kinetic energy of the ball at Position (P2) will be

$K_{2} = \frac{1}{2} m {v}^{2} \\ = > K_{2} = \frac{1}{ \cancel{2}} \times 5 \times 10 \times \cancel{10} \\ = > K_{2} = 250J............(ii)$

Work done by the gravity from position(P1) to position (P2) will be

$W_{g} = mgh$

$= > W_{g} = 5 \times 10 \times 20 \\ = > W_{g} = 1000J...........(iii)$

Finally

Denoting the work done by the resistance as

$W_{R,1 - > 2}$

and making use of eq. (i),(ii) and (iii) in work kinetic energy theorem

We have

$= > W_{1->2} = K_{2} - K_{1} \\ = > W_{g,1->2} + W_{R,1->2} = K_{2} - K_{1} \\ = > W_{g,1->2} = 250 - 1000 \\ = > \boxed{W_{g,1->2} = - 750J}$

Note-

In the attachment we have the free body diagram of the ball at some intermediate positions.