Question

A body of mass 10 kg released from the top of a tower which acquires a velocity of 10 m/s after falling through a distance of 20 m . Calculate the work done by the drag force of the air on the body ?

(Take g =10 m/s^2)

Answer 1

Answer:

1.5 kJ

Explanation:

Net acceleration of body is

aₙₑₜ = (v² - u²)/(2S)

= (10² - 0²)/(2 × 20)

= 2.5 m/s²

If a is the acceleration due to drag force

aₙₑₜ = g - a

2.5 m/s² = 10 m/s² - a

a = 7.5 m/s²

Drag force = ma

= 10 kg × 7.5 m/s²

= 75 kg m/s²

= 75 N

Work done by drag force

W = Force × Displacement

= 75 N × 20 m

= 1500 N m

= 1500 J

= 1.5 kJ