Find the Force applied to a rock by a slingshot

[jesse james]

Homework Statement

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

Homework Equations

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

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

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 don't 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

 

[haruspex]

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

No, that's the PE stored in the spring. The force at extension x is kx.

tried to solve for K using 1/2kx^2=1/2mv^2

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 F_d then -∫F(x).dx = -∫F_d.dx. For average over time, use momentum. The momentum imparted is ∫F(t).dt; if the average over time is F_t then ∫F(t).dt = ∫F_t.dt.

 

[jesse james]

Nice thanks for your reply...

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.

 

[haruspex]

that gave me the answer of 72 which is exactly double the answer that the teacher wants

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?

 

[Nickie]

I would approach this problem by first breaking it down into its fundamental principles. In this case, we are dealing with the concept of work and energy, specifically the work done by the slingshot to launch the rock.

The work done by the slingshot is equal to the change in kinetic energy of the rock. We can use the formula W = ΔKE = 1/2 * m * (vf^2 - vi^2) to calculate this. In this case, the initial velocity (vi) is 0, so the formula simplifies to W = 1/2 * m * vf^2.

We know the mass of the rock is 50 grams, or 0.05 kg, and the final velocity (vf) is 24 m/s. Plugging these values into the formula, we get W = 1/2 * 0.05 kg * (24 m/s)^2 = 14.4 J.

Now, we can use the definition of work (W = F * d) to find the average force applied to the rock. The distance (d) in this case is the distance the slingshot is stretched, which is 0.4 meters. So, we can rearrange the formula to solve for force, giving us F = W / d = 14.4 J / 0.4 m = 36 N.

Therefore, the average force applied to the rock by the slingshot is 36 N, which is the correct answer. It is important to approach problems like this by understanding the fundamental principles involved and using the appropriate formulas to solve for the desired quantity.