Uni physics and conserving energy

[Starlight]

Hey can anyone help me out with this problem

(I.e. question iii, i've put the answers for the other qs up)

An 81 kg student is launched from a bridge by his friends, some 50m above the river surface. Fortunately, he is attahed to a 30m bungee cord of spring constant 270 N/M

(i) What is the equilibrium length of the bungee cord (incl force of gravity?)

Answer was 30 + 2.94 m
= 32.9 m

(ii) Find velocity he would have if he reahed the equilibrium position and fell in free flal

Answer

1/2mv^2= mgh

1/2v^= 9.8m/s^2* 32.9 m

etc.

(iii) The student has a velocity of 23m/s at equilibrium. Find the maximum extension of the spring from its equilibrium position

Can you guys help me out with this one? (answer says 45.5m), he doesnt hit the water

[availn]

He stop falling, when all of his kinetic energy downwards is converted into spring potential energy.

0.5mv² = 0.5kx²
0.5·81·23² = 0.5·270·x²
x = 12.6m

This is how much it extends past the equilibrium point, which you found to be 32.9m. So he ends up going down 45.5m, though this isn't the maximum extension of the spring from its equilibrium, might be a mistake in the question.

[Starlight]

He stop falling, when all of his kinetic energy downwards is converted into spring potential energy.

0.5mv² = 0.5kx²
0.5·81·23² = 0.5·270·x²
x = 12.6m

This is how much it extends past the equilibrium point, which you found to be 32.9m. So he ends up going down 45.5m, though this isn't the maximum extension of the spring from its equilibrium, might be a mistake in the question.

How come we don't include gravitational potential energy? If he is above a height?

[alondouek]

Ooh, maybe I can answer this! (Don't assume I'm correct though, not much of a physicist)

Prior to jumping, when he is stationary at the top of the bridge, he has roughly 100% gravitational potential energy. This decreases as he falls, with the KE increasing proportionally to the PE_grav decrease. Then, after the equilibrium point, the KE is converted to spring potential energy as per availn's answer.

[availn]

How come we don't include gravitational potential energy? If he is above a height?

No idea, this was the only way I could get to the answer given.

[availn]

Actually, I think it's because we already take into account gravitational potential energy energy when we change the equilibrium point, so that when we do measurements from the equilibrium point, we don't need to consider it. For example, if we were to calculate everything relative to the spring's unstretched length, it would be like this:

0.5mv² + 0.5k(2.94)² + mg(x-2.94) = 0.5kx²

[Alwin]

Actually, I think it's because we already take into account gravitational potential energy energy when we change the equilibrium point, so that when we do measurements from the equilibrium point, we don't need to consider it. For example, if we were to calculate everything relative to the spring's unstretched length, it would be like this:

0.5mv² + 0.5k(2.94)² + mg(x-2.94) = 0.5kx²

I was thinking along the same lines. The solution yeilds x= -9.6576 (reject obviously) or x = 15.538.
Clearly 30+15.538=45.538m the correct answer.
It is also consistent with availn's original post using the stretched length as reference point, (unrounded) 12.598+32.94=45.538m.

[SocialRhubarb]

where x is the displacement from the equilibrium position.







Fall distance = equilibirum distance + x = 30 + 2.94 + 12.60 = 45.54

Everything actually cancelled really nicely.

[Starlight]

Thanks heaps everyone!