Determine the speed of the rock as it leaves the spring.

[HumanPerson]

Here is the question:
A spring is compressed 20cm. When it is released, it exerts an average force of 57.2 N on a 7.3 kg rock, shooting it across a frictionless floor. Determine the speed of the rock as it leaves the spring.

Homework Equations

W=change in energy
Work energy theorem
W=Fd

The Attempt at a Solution

W=Fd
=(0.20cm)(57.2N)
=11.44J

W= change in EP elastic
=1/2kx^2
Rearrange formula to solve for k (spring constant)
=10.69579357 N/m

 

[mrnike992]

I think you're going the wrong direction with this. You have the force applied and the mass affected, so you can solve for the average acceleration of the rock (F=ma, a=7.85616438 m/s²). With the acceleration and Δx (20 cm or .2 m), you can apply the formula v_f² = v_o² + 2(a)(Δx). Since v_o is 0, 2 times a times Δx = your final velocity squared.

 

[haruspex]

A spring is compressed 20cm. When it is released, it exerts an average force of 57.2 N

This is awkward.
Unless stated otherwise (e.g. as "average force with respect to distance") average force means the average over time. $F_{avg} = \frac{\int F.dt}{\int dt} = \frac {\Delta p}{\Delta t}$, where p is momentum. On that basis, you need to use the fact that the spring will expand according to simple harmonic motion.
However, it's 10 to 1 that the problem setter has screwed up and expects you to take the average force as being an average over distance: $F_{distance avg} = \frac{\int F.ds}{\int ds} = \frac {\Delta E}{\Delta s}$, so I'll answer below on that assumption.

W=Fd
=(0.20cm)(57.2N)
=11.44J

So far so good, but you no longer care about the spring. You know how much work has been done on the rock. Deduce its speed from that.

 

[mrnike992]

These give you the same answer... Work in = Kinetic Energy Final, right?

 

[NascentOxygen]

W= change in EP elastic
=1/2kx^2
Rearrange formula to solve for k (spring constant)
=10.69579357 N/m

This formula is not useful to you here, because you are told the spring equation is not that of a normal spring. In a normal spring, F=kx. But in this exercise, you are told to use F=constant, independent of x.

 

[NascentOxygen]

These give you the same answer... Work in = Kinetic Energy Final, right?

That should work.

 

[haruspex]

These give you the same answer... Work in = Kinetic Energy Final, right?

What give the same answer? Are you saying that both definitions of average force give the same answer?
This is annoying.. I posted a long response to this last night and it has disappeared. Maybe some site monitor mistakenly thought I was providing you with a complete answer to the OP, so I'm not going to try typing it all again.
The upshot was that I demonstrated that using the correct definition of average force gives a different answer The ratio between the answers is sqrt(pi/4).

 

[mrnike992]

I solved it the way I showed in my original post and with the equation (Work = kinetic energy final) and both most certainly gave me the exact same answer, accurate to like twelve or fifteen sig-figs, however many my calculator allowed.

 

[mrnike992]

using the correct definition of average force gives a different answer

Okay, I think I see what you're saying. But this is a high school physics question, I'm sure the question is implying a force over a distance, so it would be an Energy problem, as indicated by the formula for potential energy in a spring. It does say to find the speed as it leaves the spring, and the spring was compressed .2m, so I believe it would be indicating that the average force over the distance from the initial position to the end of the spring, .2m away.

 

[haruspex]

Okay, I think I see what you're saying. But this is a high school physics question, I'm sure the question is implying a force over a distance, so it would be an Energy problem, as indicated by the formula for potential energy in a spring. It does say to find the speed as it leaves the spring, and the spring was compressed .2m, so I believe it would be indicating that the average force over the distance from the initial position to the end of the spring, .2m away.

I agree that this is almost certainly what the problem setter intends here, but I hate the thought of students being taught that this is a valid interpretation of average force. It leads to the bizarre situation that mass * average acceleration does not always equal average force.