Question

A skydiver of mass m jumps from a hot air balloon and falls a distance d before reaching a terminal velocity of magnitude v . Assume that the magnitude of theacceleration due to gravity is g .

-What is the work (Wd) done on the skydiver, over the distance , by the drag force of the air?

-Find the power (P d) supplied by the drag force after the skydiver has reached terminal velocity v.

Answer 1

Concepts and reason

The concepts required to solve this question are the energy, work done, force, and the power.

Initially, find the expression for kinetic energy and potential energy of a sky driver. Then using the expression for both the kinetic energy and the potential energy, find the expression of the work done on the skydiver in the presence of air resistance.

Finally, the power can be calculated using the force which is equal to the weight of the skydiver and the velocity.

Fundamentals

The expression for the kinetic energy is,

$KE = \frac{1}{2}m{v^2}$

Here, KE is the kinetic energy, m is the mass, and v is the velocity.

The expression of the potential energy is,

$PE = mgh$

Here, PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.

The force can be expressed as the product of the mass and the acceleration. The expression of the force is,

$F = ma$

Here, F is the force, m is the mass, and a is the acceleration.

The expression of the power is,

$\begin{array}{c}\\P = F \cdot v\\\\ = Fv\cos \theta \\\end{array}$

Here, P is the power, F is the force, v is the velocity, and $\theta$ is the angle between the force vector and the velocity vector.

(a)

The skydiver jumps with velocity v from the hot air balloon.

The kinetic energy is,

$KE = \frac{1}{2}m{v^2}$

Here, m is the mass of the skydiver and v is the velocity.

The skydiver jumps from the balloon and falls a distance d.

The potential energy of the sky driver is,

$PE = mgd$

The work done ${W_{\rm{d}}}$ by the air resistance is,

${W_{\rm{d}}} = KE - PE$

Substitute $\frac{1}{2}m{v^2}$ for KE and $mgd$for PE.

${W_{\rm{d}}} = \frac{1}{2}m{v^2} - mgd$

(b)

The force is due to the weight of the skydiver.

The force is,

$F = mg$

Here, g is the acceleration due to gravity.

The power ${P_{\rm{d}}}$ supplied by the drag force after the skydiver reached the terminal velocity is,

$\begin{array}{c}\\{P_{\rm{d}}} = F \cdot v\\\\ = Fv\cos \theta \\\end{array}$

Substitute $mg$ for F and $180^\circ$ for $\theta$.

$\begin{array}{c}\\P = F \cdot v\\\\ = \left( {mg} \right)\left( v \right)\cos \left( {180^\circ } \right)\\\\P = - mgv\\\end{array}$

Ans: Part a

The work done on the skydiver by the drag of the force of the air is: $W = \frac{1}{2}m{v^2} - mgd$

Part b

The power supplied by the drag force is $- mgv$.