Chapter 6, Problem 79GP

To determine

a. The work done by the snow to bring the pilot to rest.

b. The average force exerted by the snow on the pilot.

c. The work done on him by air resistance.

Answer to Problem 79GP

Solution:

a. The work done by the snow on the pilot is approximately $-90kJ$ .

b. The average force the now exerts on the pilot is approximately $81kN$ .

c. The work done by air resistance on the pilot is approximately $-4.08\times {10}^5\text{J}$ .

Explanation of Solution

Given:

The mass of the pilot is $m=88\text{kg}$ .

The height from which he fell is $h=370\text{m}$ .

The depth on the snow due to his fall is $h_s=1.1\text{m}$ .

His impact velocity on the snow is $v=45\frac{\text{m}}{\text{s}}$ .

Formula Used:

Use the energy conservation formula:

Initial energy =Final energy

To calculate the average force, consider the formula:

work = force × displacement

Calculation:

a

Since the snow acts in a direction opposite to that of the pilot’s motion, hence the work done by it will be negative.

Total work done by the snow will be the summation of kinetic energy of the pilot and gravitational potential energy of the snow.

The work done by the snow,

  $\begin{array}{l}{W}_s=-\left(\frac{1}{2}m{v}^2+mg{h}_s\right)\\ W_s=-\left(\frac{1}{2}\times 88\times {45}^2+88\times 9.8\times 1.1\right)\text{J}\\ W_s=-90050\text{J}\\ W_s=-90\text{kJ}\end{array}$

b

Since the snow applies a force on the pilot in a direction opposite to the motion of the pilot, the work done by the snow is negative,

If the average force is $F_{av}$ , then

  $\begin{array}{l}{W}_s=-F_{av}{h}_s\\ -90050=-F_{av}\times 1.1\\ F_{av}=81,863\text{N}\\ F_{av}=81\text{}\text{kN}\end{array}$

c

The work done by air resistance on the pilot for the whole path,

  $\begin{array}{l}{W}_a=-\left(\frac{1}{2}m{v}^2+mgh\right)\\ W_a=-\left(\frac{1}{2}\times 88\times {45}^2+88\times 9.8\times 370\right)\text{J}\\ W_a=-4.08\times {10}^5\text{J}\end{array}$

Conclusion: a. The work done by the snow on the pilot is approximately $-90kJ$ .

b. The average force the now exerts on the pilot is approximately $81kN$ .

c. The work done by air resistance on the pilot is approximately $-4.08\times {10}^5\text{J}$ .