How Fast Does a Stone Shoot from a Rubber-Band Slingshot?

AI Thread Summary
A rubber-band slingshot launches a 25g stone, with a force of 27 N applied over a distance of 0.15 m, resulting in an energy of 4.05 J. The user attempts to calculate the initial speed of the stone using the kinetic energy formula but initially arrives at an incorrect speed of 5.69 m/s. The confusion arises from the conversion of mass from grams to kilograms and the application of energy formulas. After double-checking the mass conversion, the user realizes the mistake and acknowledges the need for correction. Accurate calculations are essential for determining the correct initial speed of the stone.
XxphysicsxX
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the problem is:
A rubber-band slingshot shoots a 25-gstone. What is the initial speed of the stone if the rubberband is drawnback 0.15m with aforce of 27 N?
m= 25g
d=0.15 m
f= 27 N
v(initial) = ?


Homework Equations


Ep = (1/2)Fx
Ek=(1/2) mv^2


The Attempt at a Solution


so far, I turned the forceto energy: (27N)(0.15m)= 4.05J
and then solved for v, using Ek= (1/2)mv^2
and I get v=5.69
Im so confused, this answer deffinetly does not seem right..
Im just not sure what to do with the given force here!, do I turn it into work ? (w=fd)
I've been stuck on this question for two days now,i just don't understand how any of the givens can tie in with the energy formulas..?
 
Last edited:
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Your method was correct, however your calculation for the value of v is incorrect, as it is too low. You should get a larger answer.
 
Well, I am using the exact numbers given ,just converting grams to kg, ...
Ek= (1/2)mv^2
4.05J= (1/2)(0.25)V^2
4.05J/0.125 =V^2
V=5.69

Where am I making a mistake?? :s
 
XxphysicsxX said:
Well, I am using the exact numbers given ,just converting grams to kg, ...
Ek= (1/2)mv^2
4.05J= (1/2)(0.25)V^2
4.05J/0.125 =V^2
V=5.69

Where am I making a mistake?? :s

Double check the mass.
 
Woww, I feel stupid now, haha,

Thankyou!:biggrin:!
 
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