Question

A ball of mass 5 kg is dropped vertically through a height of 20m and it gains a speed of 10 metre per second after 20. Find the work done by gravitational force and air friction

Answer 1

Explanation:

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

Answer 2

Answer:

-750J

Explanation:

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.

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

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

Finally

Denoting the work done by the resistance as

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

We have

Note-

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