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. $14000 \,J$

Answer 1

Answer: (B)

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
$=\frac{1}{2} \times 70\left(20^{2}-0^{2}\right)$
$\Rightarrow$ Work done by air drag
$=14000-35000=-21000\, J$