Question

A 62 kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behaves like a simple spring. (a) Calculate how much it would stretch if the same person were lying in it. (b) How much would it stretch if the person jumped from 38 m?

Short answer

(a) If person was lying on it, then it would stretch by\(4.6 \times {10^{ - 2}}\;{\rm{m}}\).

(b) Fire net would stretch\(1.9\;{\rm{m}}\)down if the person jumped from\(38\;{\rm{m}}\).

Step-by-step solution

Concept of elastic potential energy

In this problem, the gravitational potential energy initially is converted to elastic potential energy. Use the conservation of energy principle to find the stretched distance.

Given data

The window height from the fire net is\({y_1} = 20.0\;{\rm{m}}\).

The downward vertical displacement is\({y_2} = - 1.4\;{\rm{m}}\).

The mass of the person is\(m = 62\;{\rm{kg}}\).

The height from which the person jumps is\({y_1}' = 38\;{\rm{m}}\).

Calculation of the spring constant 

The gravitational potential energy is calculated as:

\({E_{\rm{g}}} = mgy\)

Here,\(m\)is the mass of the particle attached and\(g\)is the acceleration due to gravity.

Elastic potential energy counterbalances the gravitational potential energy above the fire net and below it.

\(\begin{aligned}{c}mg{y_1} &= mg{y_2} + \frac{1}{2}ky_2^2\\k &= 2mg\frac{{\left( {{y_1} - {y_2}} \right)}}{{y_2^2}}\\k &= 2\left( {62\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\frac{{\left( {20\;{\rm{m}} - \left( { - 1.4\;{\rm{m}}} \right)} \right)}}{{{{\left( { - 1.4\;{\rm{m}}} \right)}^2}}}\\k &= 1.3268 \times {10^4}\;{\rm{N/m}}\end{aligned}\)

Here, k is the spring constant.

Calculation of vertical displacement due to its own weight

If person was lying on the fire net, then the only forces acting would be its weight downwards and upwards reaction force due to fire net.

The vertical displacement due to stretch would be:

\(\begin{aligned}{c}F = k\left| y \right|\\mg = k\left| y \right|\\\left| y \right| = \frac{{mg}}{k}\end{aligned}\)

Substitute the known values in the above relation to find the vertical displacement.

\(\begin{aligned}{c}\left| y \right| &= \frac{{\left( {62\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{1.3268 \times {{10}^4}\;{\rm{N/m}}}}\\ &= 4.6 \times {10^{ - 2}}\;{\rm{m}}\end{aligned}\)

Hence, the vertical stretch due to the weight of the person is \(4.6 \times {10^{ - 2}}\;{\rm{m}}\).

Calculation of vertical displacement due to vertical force

Now, person jumps from\(38\;{\rm{m}}\)on the fire net. Here kinetic energy of the upwards and downwards motion will cancel out each other. So, only potential energy will be effective.

The vertical displacement due to stretch would be:

\(\begin{aligned}{c}mg{y_1}' &= mg{y_2} + \frac{1}{2}ky_2^2\\y_2^2 + \frac{{2mg}}{k}{y_2} - \frac{{2mg}}{k}{y_1}' &= 0\end{aligned}\)

Substitute the known values in the above relation to find the vertical displacement.

\(\begin{aligned}{c}y_2^2 + \frac{{2\left( {62\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{\left( {1.3268 \times {{10}^4}\;{\rm{N/m}}} \right)}}{y_2} - \frac{{2\left( {62\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{\left( {1.3268 \times {{10}^4}\;{\rm{N/m}}} \right)}}\left( {38\;{\rm{m}}} \right) &= 0\\y_2^2 + 0.0916{y_2} - 3.4804 &= 0\end{aligned}\)

On further solving the above relation.

\({y_2} = - 1.9119\;{\rm{m,}}\;1.8204\;{\rm{m}}\)

Hence, the vertical stretch due to the weight of the person is approximately,\(1.8204\;{\rm{m}}\).