Question

What is the height of the platform?

A 45.2 kg circus performer jumps from a platform into a safety net below. The net, which has a force constant of 2.8 × 104 N/m, is stretched by 1.19 m. If the unstretched net is positioned 0.65 m above the ground, what is the height of the platform? Ignore the effects of air resistance

Answer 1

I tried this BUT I suspect there is something wrong...If the net is placed at `0.65m` and stretches `1.19m` when the guy falls in it...well, he will hit the ground!

Explanation:

I would use conservation of Mechanical Energy between the initial jumping position A and the final position B inside the safety net at maximum extension.

Answer 2

A 45.2 kg circus performer jumps from a platform into a safety net below. The net, which has a force constant of 2.8 × 104 N/m, is stretched by 1.19 m. If the unstretched net is positioned 0.65 m above the ground, what is the height of the platform? Ignore the effects of air resistance

Let's use energy! If you haven't learned it yet, this can be done with forces, it's just much more work in my opinion.

The initial energy of the performer is just `mgh`, where the height will be measured relative the ground.

The final energy is `1/2 kx^2 + mgh_f` where the final height is also measured relative the ground. This is given: If the net is initially 0.65 m off the ground and it is dipped down by 1.19 m, the final height is -0.54 m. You may be thinking "Negative??? How is it negative?". Fear not, young physicist. The problem gives us this strange situation. I believe the problem requires a bit of a hole in the ground, which is actually not particularly strange theoretically.

From there, we can equate energies and get the following:
`mgh = 1/2 kx^2 + mgh_f rightarrow h = (kx^2)/(2mg) + h_f`
so, plugging in numbers
`h = (2.8 * 10^4 N/m * (1.19\ m)^2)/(2 * 45.2\ kg * 9.8 N/(kg)) = 44.8\ m`

which is our initial height relative the ground.