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. The work done by the viscous
air drag is
a. 21000
b . -21000
c. -14000
d . 14000

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Answer 1

Concept Introduction:-

Work is described as the result of how much the displacement is (d) and the force's component acting in the displacement direction.

Given Information:-

We have been given that 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.

To Find:-

We have to find that The work done by the viscous air drag is

Solution:-

According to the problem

$W_{g}+W_{v}=\Delta K E$

$=\frac{1}{2} \times 70\left(20^{2}-0^{2}\right)\\=\frac{1}{2} \times 70\times 400\\ =\frac{10 \times 400}{2}$

$W_{g}=70 \times 10 \times 50\\ W_{v}=14000-35000=-21000 \mathrm{~J}$

Final Answer:-

The correct answer is $-21000$ J.

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