# Homework Help: Need help finding velocity of spring launched object

1. Jul 10, 2012

1. The problem statement, all variables and given/known data
A toy company hires you to construct a device that uses a small spring to launch 15 g foam pellets. The spring requires an average force of 5.70 N to compress it 1.30 cm. Determine the speed of one of the foam pellets as it leaves the spring.

2. Relevant equations
Ep=-Ek

3. The attempt at a solution

EP=-EK
EP=MGH
EP.015(9.81)(.013)
EP=.0019
EK=-.0019

V=SQRT(EK2/M)
V=SQRT(-.0019)(2)/.015

I know this is wrong as you cannot square a negative number. I dont understand where the force given in this equation comes in to play either.

2. Jul 10, 2012

### Staff: Mentor

You are right to try to use energy conservation for this problem. But there is no place for gravitational potential energy E=mgh here.

Instead, you need to figure out the potential energy stored in the spring, and use that to figure out the initial velocity as the mass leaves the spring flying away.

Normally you would use the equation for the potential energy of a compressed spring. Can you write that out for us? It involves the spring constant and the distance the spring is compressed.

However, in this problem you are not explicitly given the spring constant. Instead, you are given an average force and the distance of compression. What energy equation can you think of that would involve those two quantities, that would help you calculate the initial energy stored in the spring?

3. Jul 10, 2012

### TSny

I don't think gravitational potential energy is relevant in this problem. The type of potential energy you are dealing with here is elastic potential energy stored in the spring. Do you know a formula for that type of potential energy?

The equation Ep = -Ek requires careful interpretation. The general idea is conservation of total energy. So, you start with all the energy stored as potential energy in the spring and no kinetic energy. As the spring decompresses, the potential energy of the spring decreases while the kinetic energy of the pellet increases. So, what you want to say is that the loss of potential energy equals the gain in kinetic energy. Or better, the change in potential energy equals the opposite (negative) of the change in kinetic energy: ΔEp = -ΔEk.

In terms of initial (i) and final (f) values of energy, this would read

Epf - Epi = -(Ekf - Eki).

Think about what you would substitute for each of these quantities. If you're careful, the signs will all work out correctly.

[Note added: I see berkeman beat me to the punch. Hope I'm not being too redundant.]

4. Jul 10, 2012

### azizlwl

If the pellet is launch at a certain angle to the horizontal, potential energy is relevant.

Last edited: Jul 10, 2012
5. Jul 10, 2012

Would it be hookes law, rearranged to give me the spring constant?

How would I use that to find the initial energy stored in the spring?

I think I could use some clarification on how the energy of a spring works. My understanding is that force being applied to the spring is being stored as potential energy. Force being applied is kinetic energy? What is the energy in the spring when the spring is not compressed?
Would that be the mechanical energy?

6. Jul 10, 2012

Also, how would I determine the tension at this point. Would the equation T=mv^2/r -mgsin(theta) be relevant?

7. Jul 10, 2012

### TSny

The potential energy stored in the spring is equal to the work required to compress it. Since you are given the distance of compression and the average force used to compress it, you can find the work. So you don't need Hooke's law, the spring constant, or the formula for the potential energy of the spring for this problem.

Azizlwl, I agree the problem is not clear concerning how the pellet is launched. But gravitational potential energy is a relatively small effect. The force of gravity on the pellet is mg = .147 N while the average force of the spring is 5.70 N. So, the change in gravitation potential energy is, at most, only about 2.6% of the change in potential energy of the spring. The problem doesn't say if the pellet is being launched vertically. If it's being launched at an angle, then that would decrease the change in potential energy of gravity and if it's being launched horizontally then there would be no change in gravitational potential energy.

8. Jul 10, 2012

So the potential energy in the spring after compression is 1/2Fx? I am lost as to what to do next lol sorry for incompetency.

9. Jul 11, 2012

### TSny

The work is just $\bar{F}$x where $\bar{F}$ is the average force exerted while compressing the spring. This is given to be 5.7 N.

While compressing the spring, your force would start at 0 and end at some final force, Ff, where Ff = kx. The average force would be $\bar{F}$ = (1/2)(0 + Ff) = (1/2)(0 + kx) = (1/2)kx. So the work is $\bar{F}$x =[(1/2)kx]x=(1/2)kx2. This gives the formula for the potential energy of a spring Ep = (1/2)kx2.

But, in this problem you are given the average force, so the potential energy of the spring is just $\bar{F}$x which you can calculate from the numbers given in the problem.

When the pellet is fired, the stored potential energy is converted into kinetic energy. So, the final kinetic energy of the pellet is a positive number equal to the initially stored potential energy.

10. Jul 11, 2012

Ok so the intially stored Ep is Fx and which makes the Ek the same as Ep when the pellet is fired. So to find the velocity would I take the 1/2 out of the 1/2mv^2 equation? Or would it remain the same?

11. Jul 11, 2012

### TSny

The 1/2 in the kinetic energy would remain.

12. Jul 11, 2012

### CWatters

PE in spring = Average Force * Displacement
KE in pellet = 0.5 * Mass * Velocity2

Equate.