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

]​

Answer 1

Given : 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.

To Find :  work done by air friction

Solution:

object of mass 10 kg is dropped from a tower of height 150 m

h = 150 m

v = 0

PE = mgh = 10 * 10 * 150 = 15000  J

KE = (1/2)mv² = 0  J

TE = PE + KE = 15000 + 0 = 15000 J

The body strikes the ground with velocity 40 ms–1.

v = 40 m/s

h = 0

PE = mgh =  0  J

KE = (1/2)mv² = (1/2)10(40)² =   8000J

TE = PE + KE = 0 + 8000   = 8000 J

work done by air friction  =   Change in Energy

= 8000  - 15000

= -7000 J

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Answer 2

Given:

An object of mass 10 kg is dropped from a tower of height 150 m. The body strikes the ground with velocity 40 m/s.

To find:

Work done by air friction ?

Calculation:

In these kinds of questions, it is best to apply WORK-ENERGY THEOREM which states that:

• The net work done by all the forces acting on a body will be equal to the change in kinetic energy of the body.

So, applying Work-Energy Theorem:

$\therefore \: W_{gravity} + W_{air \: friction} = \Delta KE$

$\implies\: mgh+ W_{air \: friction} = \dfrac{1}{2} m {v}^{2} - \dfrac{1}{2} m {u}^{2}$

$\implies\: mgh+ W_{air \: friction} = \dfrac{1}{2} m {v}^{2} - \dfrac{1}{2} m {(0)}^{2}$

$\implies\: mgh+ W_{air \: friction} = \dfrac{1}{2} m {v}^{2}$

$\implies\: (10 \times 150 \times 10)+ W_{air \: friction} = \dfrac{1}{2} \times 10 \times {(40)}^{2}$

$\implies\: 15000+ W_{air \: friction} = 8000$

$\implies\: W_{air \: friction} = - 7000 \: joule$

So, work done by air resistance is -7000 Joule.