Question

A 92.0-kg skydiver falls straight downward with an open parachute through a vertical height of 325 m. The skydiver’s velocity remains constant. What is the work done by the nonconservative force of air resistance, which is the only nonconservative force acting?

Answer 1

Answer:

The work done is $293.0kJ$

Explanation:

Data

• Mass:   $m=92.0kg$

• height:    $h=325m$

• acceleration due to gravity:  $g=9.81m/s^{2}$

Non conservative force act

The work is:

$W=$Δ$KE$+Δ$PE$

KE and PE means kinetic energy and potential energy respectively.

The change in kinetic energy is 0, since the skydiver’s velocity remains constant.

Therefore

$W=0+$Δ$PE$

$W=$Δ$PE$

$W=mgh$

substitute the values

$W=92.0kg*9.81m/s^{2} *325m\\W=293020J\\W=293.0kJ$