A person of mass 70 kg jumps from a stationary helicopter with the parachute open. As he falls through 50 m height, he gains a speed of 20 ms -1. The work done by the viscous air drag is

Question

A person of mass 70 kg jumps from a stationary helicopter with the parachute open. As he falls through 50 m height, he gains a speed of 20 ms -1. The work done by the viscous air drag is (A) 21000 J (B) -21000 J (C) -14000 J (D) 14000 J

Tardigrade Admin · Accepted Answer

From work-energy theorem, net work done by all forces (internal and external) = change in kinetic energy.

\Rightarrow

Work done by gravity + work done by air drag

= \frac{1}{2} \times 70 (2 0^{2} - 0^{2})

\Rightarrow

Work done by air drag

= 14000 - 35000 = - 21000 J

Q. A person of mass $70 k g$ jumps from a stationary helicopter with the parachute open. As he falls through $50 m$ height, he gains a speed of $20 m s^{- 1}$ . The work done by the viscous air drag is

$21000 J$

$- 21000 J$

$- 14000 J$

$14000 J$

From work-energy theorem, net work done by all forces (internal and external) = change in kinetic energy.
$\Rightarrow$ Work done by gravity + work done by air drag
$= \frac{1}{2} \times 70 (2 0^{2} - 0^{2})$
$\Rightarrow$ Work done by air drag
$= 14000 - 35000 = - 21000 J$

Q. A person of mass 70kg jumps from a stationary helicopter with the parachute open. As he falls through 50m height, he gains a speed of 20ms−1. The work done by the viscous air drag is

21000J

−21000J

−14000J

14000J