With what speed does the stone leave the slingshot?

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SUMMARY

The discussion focuses on calculating the potential energy stored in a slingshot and the speed at which a stone leaves it. The force required to stretch each rubber band is 15 N for 1.0 cm, leading to a spring constant (k) of 1500 N/m when converted to meters. The potential energy (PE) in the two bands when a 47g stone is pulled back 12 cm is calculated using the formula PE = 0.5 * k * x^2, resulting in a total PE of 1080 J. The speed of the stone upon release can be determined using energy conservation principles.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with potential energy calculations
  • Basic knowledge of energy conservation principles
  • Ability to convert units (e.g., cm to m)
NEXT STEPS
  • Calculate the speed of the stone using the formula v = sqrt(2 * PE / m)
  • Explore the implications of using multiple springs in parallel
  • Investigate the effects of different materials on spring constants
  • Learn about energy transfer in elastic systems
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Physics students, engineering students, and hobbyists interested in mechanics and energy transfer in elastic systems will benefit from this discussion.

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A slingshot consists of a light leather cup containing a stone that is pulled back against two parallel rubber bands. It takes a force of 15 N to stretch either one of the bands 1.0 cm. A)What is the potential energy in the two bands together when a 47g stone is placed in the cup and pulled back.12 m from the equilibrium position? B) With what speed does the stone leave the slingshot? I'm not sure where to start!
 
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you have to consider the rubber band has happy Hookian springs. I.e. such that F = kx.

By "It takes a force of 15 N to stretch either one of the bands 1.0 cm", what they mean is that if you exert a force of 15N on one of the rubber band, you're not going to be able to stretch it farther than 1cm. At this point, equilibrium will have been attained. this means that at x = 1cm, the force of the rubber band is of 15N too. From this, you get k = 15 [N/cm].

You go from there.
 
Last edited:
so... then would 7.5 be the correctanswer to part A?
 
Nope. How did you get that answer?

(Btw - I fixed some dimension errors in my first post. I replaced all meters with centimeters)
 
Last edited:
I used the equation PE=.5kx^2and put 15 infor k and 1 in for x.
 
That's the right idea, but pay more attention.

The actual problem talks about a rock that is pulled back.12 m (12 cm, not 1cm). Moreover, 15 [N/m] is the spring constant of ONE rubber band. You got two.
 

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