Question

If you're launching a 15g marble using a spring, and the spring requires an average force of 5.70 N to compress it to 1.30 cm, how would one find the speed of the marble as it leaves the spring?

Answer 1

You don't need to post the numerical answer to the question, just the process/equations used. Thanks

Answer 2

I'm not so good with physics, or anything else at that matter. First though, what is the spring force formula, and constant (k)?

Answer 3

No spring constant is given, and the spring formula for Hooke's Law is \[E _{p} = \left(\begin{matrix}1 \\ 2\end{matrix}\right) k x ^{2}\]

Answer 4

I'm just tiding you over until either someone better comes or you get it yourself, but what do the variables mean?

Answer 5

Or F = kx

Answer 6

Where:
F is force exerted by the spring (N)
Ep is the energy (J)
k is the spring constant (N/m)
x is the displacement of the force end of the spring (N/m)

Answer 7

What's the m in the n/m?

Answer 8

I think I might have figured it out, just one second

Answer 9

Yeah I got it, thanks though!

Answer 10

Wonderful. Glad I 'helped'.

Answer 11

We can just plug the average spring force into the work equation. \[W = F_{Avg} \cdot d\]where d will be the displacement of the spring.

Once we know work, we can use the Work-Energy Theorem to find the speed of the marble as it leaves the spring. Recall that the Work-Energy Theorem states that work done on a system equals the change in energy of that system.\[W = \Delta E = \Delta KE + \Delta PE\]We can assume that \(\Delta PE = 0\).

Therefore,\[W = \Delta KE\]
We can obtain the following\[F_{avg} \cdot d = {1 \over 2} m v^2\]where v is the velocity as te marble leaves the spring. I assumed the marble was at rest initially.

Answer 12

As an aside, we can only use\[W = F \cdot d\] when average force is given. If we knew the spring constant and displacement, we would use the following form of the work equation\[W = \int\limits F dx = {1 \over 2} kx^2\]