# How high will a stone shot from a slingshot go?

1. Jul 18, 2009

### colachuu

1. The problem statement, all variables and given/known data
(A)A rubber band slingshot shoots a 25g stone What is the intial speed of a stone if the rubber band is drawn back 0.15m with a force of 27N?
(B) How high will the stone rise if it is shot straight upward?

m=0.025kg
x=0.15m
F=27N

2. Relevant equations
Ep=1/2Fx
For this question in order to find velocity i believe one has to assume
Ep=Ek
Ek=1/2mv2
v2=2W/m

d=W/F

3. The attempt at a solution
Ep=1/2Fx
Ep=1/2(27N)(0.15m)
Ep=2.025J
Elastic potential energy is converted to kinetic energy when it is shot from the slingshot therefore:
Ep->Ek
2.025J=1/2mv2
v2=(2.025 J)(2)/(0.025kg)
v=12.73m/s

I believe that this would be the initial speed of the stone after being shot however i am having difficulties determining how high the stone would rise if it is shot straight upwards as asked in part (B)

The formula that i think would work is
d=W/F
however im not sure what W and F are especially considering gravity.
How do i determine these factors??

Thank-you for taking a look!

2. Jul 18, 2009

### RoyalCat

Where did you get that equation for $$E_p$$?

Potential energy is the energy it would be required to bring an object from point A to point B (Making sure the final velocity at point B is 0) while overcoming some preserving force.

Where did you get the factor of $$\frac{1}{2}$$ ? I think you're getting things mixed up with potential elastic energy for a spring abiding by Hooke's law.

$$E_p (y) = -\int \vec F dy$$
Now, since our force is invariant (It is constant at 27N, correct?), what will the integral be? Consider the case of gravity (Where the dependence on the height is linear!)

If we assume that the rubber-band's elastic tension is a preserving force, and that once the stone is released, whatever potential energy we "charge" the rubber-band with is transferred to the stone, then the calculations become quite simple.

At first, we charge the rubber-band with potential energy. We perform work on it in order to do so. Going by the phrasing of the question, I'd ignore the effect of gravity on this part.
I suggest you assume that the rock is mounted on the rubber-band AFTER it's "charged" with potential energy.
$$W=F\cdot d$$

If we assume energy is preserved, then the initial potential energy is equal to the kinetic energy at the point of release (Remember that the rubber-band has been pulled $$0.15m$$ below this point), and to the potential height energy at the maximum height.

Last edited: Jul 18, 2009
3. Jul 18, 2009

### JazzFusion

I don't think that the force is constant. It sounds to me as if you are expected to assume the rubber band in the slingshot acts like a spring obeying Hooke's Law. This would be a very reasonable assumption for how a slingshot behaves.

Q1: If a 27N weight were hanging on a spring under the force of gravity, and the spring stretched 0.15 cm, what would be the spring constant?

Q2: If a 27N force pulled on a spring (maybe gravity, maybe not), and the spring stretched 0.15 cm, what would be the spring constant?

Q3: If you know the spring constant and the amount the spring (or rubber band) is pulled, what is the potential energy?

Q4: If you know the potential energy, what is the final height, assuming energy is conserved?

4. Jul 20, 2009

### colachuu

Potential energy can be found simply with the force, so finding the spring constant can be slightly redundant which is why i used
Ep=1/2Fx
rather than
Ep=1/2kx2 seeing as F=kx

Therefore JazzFusion for
A3: the potential energy is
Ep=1/2(27N)(0.15m)
Ep=2.025J

So the elastic potential energy (according to Hooke's law) is 2.025J.

Will all the elastic potential energy convert to gravitational energy when it reaches max height? Meaning can i assume that:

elastic Ep=gravitational Ep
2.025J=mgh
and solve for height from there?

5. Jul 20, 2009

### queenofbabes

Yes, for this problem you can get away with using Ep = 1/2 Fx, just don't get confused over what you're really doing.

elastic Ep=gravitational Ep should be fine.

6. Jul 21, 2009

### JazzFusion

Yes, that is the point of writing it out in terms of energy - it makes this relationship transparent (and it applies in the general case of any position, any combination of x, h, v).

7. Jul 21, 2009

### vorcil

Your elastic potential equation was wrong, It's Ep = -Kx where k is your spring constant for the Rubber band and x is the displacment it's pulled back

8. Jul 22, 2009

### JazzFusion

No, his equation for Ep is correct. The Potential Energy for a spring obeying Hooke's Law is Ep = 1/2 k x2. The force is F = -kx. What he did was solve for 'k' algebraically and substitute F/x back into the equation for potential energy.