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

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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..?
 
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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.