Question

An object of mass 10 kg is dropped from a tower of height 150 m. The body strikes the ground with velocity

40 ms–1. Calculate work done by air friction. [Take g = 10 m/s2]

pls show the steps!​

Answer 1

Answer:

mass, m = 1kg

height, s = 20m

initial velocity of ball, u = 0m/s

acceleration, a = 10m/s  

2

 

Using, v  

2

 = u  

2

 + 2as

               = 0 + 2×10 ×20

               = 400

            v = 20m/s

Now, let us take upward direction as positive and downward direction as negative.

Initial momentum of the ball (before striking the ground) = m×v

                                                                                             = 1×(-20)  (velocity is downward, hence negative)

                                                                                             = -20kgm/s  

Now, when the ball strikes and rebounds, its velocity becomes +20m/s (after striking, it moves upward, hence positive velocity)

Final momentum (after striking the ground) = m×v

                                                                        = 1×(+20)

                                                                        = +20kgm/s

Change in momentum = Final momentum - Initial momentum

                                     = 20kgm/s - (-20kgm/s)

                                     = 40kgm/s in upward direction.

Explanation:

Answer 2

Given:

Mass of the object = 10 kg

Height from which the object is dropped = 150 m

Final velocity = 40 m/s

g= 10 m/s²

To find:

The work done by air friction.

Solution:

Work done = force× displacement.

The force is the weight of the object:

Therefore force = 10* 10 = 100N

Putting the values in the equation we get:

Work = 100* 150 = 15000 J or 15 kJ

Work done by air friction is 15 kJ.