Question

A 70.0 kgman jumping from a window lands in an elevated fire rescue net 11.0 mbelow the window. He momentarily stops when he has stretched the net by 1.50 m. Assuming that mechanical energy is conserved during this process and that the net functions like an ideal spring, find the elastic potential energy of the net when it is stretched by 1.50 m.

Short answer

The elastic potential energy of the net when it is stretched by 1.5 m is 8580 J.

Step-by-step solution

Given data:

Mass of man, m=70.0 kg

Distance of window above net, h=11.0 m

Net is stretched by, 1.5 m

To understand the concept:

As the man jumps from the window, which means he starts from rest, he will have zero initial kinetic energy. According to conservation of energy, the potential energy will be equal to the elastic potential energy of the net relative to where he momentarily stops.

Formula:

$PE=mgh$

Here, m is the mass, g is the acceleration due to gravity having a value $9.8m/s^2$, and h is the height.

Calculate the elastic potential energy of the net when it is stretched by :

As the man is starting from rest, he will have no initial kinetic energy.

Now he falls down from a distance of 11.0 m.

And net is stretched by 1.5 m.

So the total height becomes

$$\begin{gathered} H=\left(Dis\tan ceofwindowaboveground\right)+\left(Dis\tan ceofstretchednet\right) \\ =11.0+1.5m \\ =12.5m \end{gathered}$$

According to law of conservation of energy,

$$\begin{gathered} \left(ElasticPotentialEnergy\right)=(Potentialenergyat12.5m) \\ =mgH \\ =70\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 12.5\mathrm{m} \\ =8580\mathrm{J} \end{gathered}$$

Hence, the elastic potential energy of the net when it is stretched by 1.5 m is 8580 J.