Find the Force applied to a rock by a slingshot

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Homework Help Overview

The discussion revolves around calculating the average force applied to a rock by a slingshot, given its mass, launch speed, and stretch distance. The problem involves concepts from mechanics, particularly energy conservation and spring force.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between spring force and energy conservation, questioning the correct formula to use for calculating average force. There is discussion about whether the average force should be calculated over distance or time, leading to different interpretations of the problem.

Discussion Status

Some participants have provided guidance on the use of energy conservation principles and the distinction between potential energy and force calculations. There is an ongoing exploration of the implications of different interpretations of "average force," but no consensus has been reached on the correct approach.

Contextual Notes

Participants express uncertainty regarding the definitions of average force in relation to distance versus time, which may affect their calculations. There is also mention of a discrepancy between calculated answers and the expected result, prompting further inquiry into the formulas used.

jesse james
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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
 
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jesse james said:
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 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.
 
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.
 
jesse james said:
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?
 

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