# Find the Force applied to a rock by a slingshot

1. Dec 7, 2012

### jesse james

1. The problem statement, all variables and given/known data

A sling shot can launch a 50gram rock at a speed of 24m/s if it is stretched .4meters before release. Find the average force applied to the rock

2. Relevant equations

Spring force = 1/2*k*x^2

1/2kx^2 = 1/2mv^2

3. The attempt at a solution

tried to solve for K using 1/2kx^2=1/2mv^2 <-- someone told me to use this formula i dont know if its right

k=180

put that in the formula for spring force Fs=1/2kx^2
gave me the answer of 14.4N

however the correct answer should be 36N

2. Dec 7, 2012

### haruspex

No, that's the PE stored in the spring. The force at extension x is kx.
That formula is correct. It expresses the conservation of energy in this scenario.
Having found k, you then know the force at any particular extension.
But I have a problem with the question. Is the average force the average over distance or the average over time? This will yield two different answers.
For average over distance, use energy again. The energy imparted is -∫F(x).dx; if the average over distance is Fd then -∫F(x).dx = -∫Fd.dx. For average over time, use momentum. The momentum imparted is ∫F(t).dt; if the average over time is Ft then ∫F(t).dt = ∫Ft.dt.

3. Dec 7, 2012

### jesse james

so I was using the wrong formula for the force of the spring once i found K...so I found K the same way as before

then instead of using the formula for Pe of the spring I used hookes formula for the force of a spring which was force=Kx

that gave me the answer of 72 which is exactly double the answer that the teacher wants ...so am i missing a 1/2 in a formula somewhere??..

im not sure if its average over distance or average over time....I am still new at physics and get confused by some of the signs and formulas.

4. Dec 7, 2012

### haruspex

Looks like it must be average over distance that's wanted. How did you calculate the 72? Isn't that the force when at extension 0.4m?