Find the Force applied to a rock by a slingshot

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SUMMARY

The discussion focuses on calculating the average force applied to a 50-gram rock launched by a slingshot at a speed of 24 m/s after being stretched 0.4 meters. The correct approach involves using Hooke's Law, where the force is calculated as F = kx, with k determined from the conservation of energy equation 1/2 kx² = 1/2 mv². The user initially miscalculated the spring constant k as 180, leading to an incorrect force of 72 N, which is double the expected average force of 36 N. The confusion arose regarding whether to calculate average force over distance or time.

PREREQUISITES
  • Understanding of Hooke's Law (F = kx)
  • Knowledge of energy conservation principles in physics
  • Familiarity with momentum concepts
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation and application of Hooke's Law in various contexts
  • Learn about energy conservation in mechanical systems
  • Explore the relationship between force, mass, and acceleration using Newton's laws
  • Investigate the differences between average force calculations over distance and time
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Students studying physics, particularly those learning about mechanics and forces, as well as educators looking for examples of common misconceptions in force calculations.

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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