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

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Homework Help Overview

The problem involves determining the initial speed of a stone shot from a rubber-band slingshot, given the mass of the stone, the distance the rubber band is drawn back, and the force applied. The context is rooted in energy conservation principles, specifically relating to elastic potential energy and kinetic energy.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss converting the force applied to energy and using energy equations to find the initial speed. There is confusion regarding the calculations and the application of the given force in relation to energy formulas.

Discussion Status

Some participants are exploring the calculations and questioning the correctness of their results. There is acknowledgment of a potential mistake in the calculation of speed, and guidance is offered to double-check the mass conversion. The discussion reflects a mix of attempts and clarifications without reaching a consensus on the solution.

Contextual Notes

Participants are working under the constraints of the problem's parameters, including the mass in grams needing conversion to kilograms, and the relationship between force, work, and energy. There is an ongoing uncertainty about how to properly apply the given values in the context of energy equations.

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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