With what speed does the stone leave the slingshot?

In summary: So the spring constant k = 30 [N/m].In summary, a slingshot is made up of a leather cup with a stone and two parallel rubber bands. It takes a force of 15 N to stretch one of the bands 1.0 cm. With a 47g stone pulled back.12 m from equilibrium, the potential energy in the two bands combined is 41.58 J. The speed of the stone leaving the slingshot can be calculated using the equation PE = .5kx^2, with k = 30 [N/m] and x = .12 m, resulting in a speed of 4.08 m/s.
  • #1
imdumb
3
0
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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  • #2
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.
 
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  • #3
so... then would 7.5 be the correctanswer to part A?
 
  • #4
Nope. How did you get that answer?

(Btw - I fixed some dimension errors in my first post. I replaced all meters with centimeters)
 
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  • #5
I used the equation PE=.5kx^2and put 15 infor k and 1 in for x.
 
  • #6
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.
 

What is the definition of speed?

Speed is a measure of how fast an object is moving, typically measured in meters per second (m/s) or kilometers per hour (km/h).

What factors affect the speed of the stone leaving the slingshot?

The speed of the stone leaving the slingshot is affected by the tension of the slingshot, the angle at which it is released, the weight and size of the stone, and any external forces such as wind resistance.

How can the speed of the stone leaving the slingshot be calculated?

The speed of the stone can be calculated using the equation v = √(2gh), where v is the speed, g is the acceleration due to gravity, and h is the height from which the stone is released. This assumes no air resistance and a horizontal release.

Does the speed of the stone change as it travels through the air?

Yes, the speed of the stone will change as it travels through the air due to the effects of air resistance. The stone will slow down as it moves through the air, and its speed will depend on its weight, shape, and the surrounding air conditions.

Can the speed of the stone leaving the slingshot be increased?

Yes, the speed of the stone can be increased by increasing the tension of the slingshot, using a heavier or smaller stone, or releasing the stone at a higher angle. However, there will be a limit to how fast the stone can travel due to external factors like air resistance and the strength of the slingshot.

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