Question

An airplane pilot fell 370 m after jumping from an aircraft without his parachute opening. He landed in a snow bank, creating a crater 1.1 m deep, but survived with only minor injuries. Assuming the pilot’s mass was 88 kg and his speed at impact was \({\bf{45}}\;{{\bf{m}} \mathord{\left/{\vphantom {{\bf{m}} {\bf{s}}}} \right.} {\bf{s}}}\), estimate: (a) the work done by the snow in bringing him to rest; (b) the average force exerted on him by the snow to stop him; and (c) the work done on him by air resistance as he fell. Model him as a particle.

Short answer

(a) The work done by the snow on the pilot is \(9.0 \times {10^4}\;{\rm{J}}\).

(b) The average force exerted by the snow on the pilot is \(8.2 \times {10^4}\;{\rm{N}}\).

(c) The work done by the air on the pilot is \(2.3 \times {10^5}\;{\rm{J}}\).

Step-by-step solution

Define the work-energy theorem

The work done on a particle to change its state is equivalent to the sum of the variation in the kinetic and potential energies of the object in a specific state.

Given information

The height is \(h = 370\;{\rm{m}}\).

The depth is \(d = 1.1\;{\rm{m}}\).

The mass of the pilot is \(m = 88\;{\rm{kg}}\).

The final velocity of the pilot is \({v_2} = 45\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).

Since the pilot starts from rest, hisinitial velocity is\({v_1} = 0\).

Calculate the work done by the snow on the pilot

(a)

The expression for the change in kinetic energy is \(\Delta KE = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2\).

The work done by gravity and the snow is equal to the change in kinetic energy. Therefore,

\(\begin{aligned}{W_{{\rm{gravity}}}} - {W_{{\rm{snow}}}} &= \Delta KE\\mgd - {W_{{\rm{snow}}}} &= \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2\\{W_{{\rm{snow}}}} &= mgd - \frac{1}{2}mv_2^2 + \frac{1}{2}mv_1^2\\{W_{{\rm{snow}}}} &= m\left( {gd - \frac{{v_2^2}}{2} + \frac{{v_1^2}}{2}} \right)\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{W_{{\rm{snow}}}} &= \left( {88\;{\rm{kg}}} \right)\left( {\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {1.1\;{\rm{m}}} \right) - \frac{{{{\left( {45\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)}^2}}}{2} + \frac{{{{\left( 0 \right)}^2}}}{2}} \right)\\{W_{{\rm{snow}}}} &= - 9.0 \times {10^4}\;{\rm{J}}\end{aligned}\)

Here, a negative sign indicates the work done by the snow is in the opposite direction to the motion of the pilot.

Thus, the work done by the snow on the pilot is \(9.0 \times {10^4}\;{\rm{J}}\).

Calculate the average force exerted by the snow on the pilot

(b)

The average force exerted by the snow can be calculated as shown below.

\(\begin{aligned}{W_{{\rm{snow}}}} &= {F_{{\rm{snow}}}}d\cos \left( {180^\circ } \right)\\{W_{{\rm{snow}}}} &= - {F_{{\rm{snow}}}}d\\{F_{{\rm{snow}}}} &= \frac{{ - {W_{{\rm{snow}}}}}}{d}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{F_{{\rm{snow}}}} &= \frac{{ - \left( {9.0 \times {{10}^4}\;{\rm{J}}} \right)}}{{\left( {1.1\;{\rm{m}}} \right)}}\\{F_{{\rm{snow}}}} &= 8.2 \times {10^4}\;{\rm{N}}\end{aligned}\)

Thus, the average force exerted by the snow on the pilot is \(8.2 \times {10^4}\;{\rm{N}}\).

Calculate the work done by the air on the pilot

(c)

The work done by the air on the pilot can be calculated by using the work-energy theorem.

\(\begin{aligned}{W_{{\rm{gravity}}}} + {W_{{\rm{air}}}} &= \Delta KE\\mgh + {W_{{\rm{air}}}} &= \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2\\{W_{{\rm{air}}}} &= m\left( {\frac{{v_2^2}}{2} - \frac{{v_1^2}}{2} - mgh} \right)\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{W_{{\rm{air}}}} &= \left( {88\;{\rm{kg}}} \right)\left( {\frac{{{{\left( {45\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)}^2}}}{2} - \frac{{{{\left( 0 \right)}^2}}}{2} - \left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {370\;{\rm{m}}} \right)} \right)\\{W_{{\rm{air}}}} &= - 2.3 \times {10^5}\;{\rm{J}}\end{aligned}\)

Here, a negative sign indicates that the air resistive force acts in theopposite direction to the movement of the pilot.

Thus, the work done by the air on the pilot is \(2.3 \times {10^5}\;{\rm{J}}\).