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If I have a slingshot and I am told that when pulled it has about $E$ joules of energy, how do I compute the height at which an object of mass $m$ would travel if the slingshot is released in the vertical direction?

I would imagine that the energy stored is potential energy, thus related to height $h$ through the following: $E = mgh$.

Is this correct or is there any other dissipating force strongly affecting the calculation?

I.e. If I have a slingshot with 1000 joules, and an object of mass 100 kilos will it travel $h = \frac{E}{m\times g} = \frac{1000}{100\times9.8} \approx 1$ meter vertically in the air?

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closed as off-topic by ACuriousMind, Kyle Kanos, HDE 226868, Qmechanic Jul 30 '15 at 18:19

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  • $\begingroup$ Make sure you specify the point above which you measure: it has to be the height above the fully extended catapult (because you do work against gravity as the rubber band contracts) $\endgroup$ – Floris Jul 30 '15 at 18:30
  • $\begingroup$ @ACuriosMind - This question is not homework related. I've seen a video on youtube about slingshots and got curios about the physics behind them. $\endgroup$ – Matteo Aug 3 '15 at 15:47
  • $\begingroup$ @Floris - I would like it to be the height from the position of rest (so in this case it would be ground). Suppose the slingshot has already been contracted and the object is in place, what happens when you release? $\endgroup$ – Matteo Aug 3 '15 at 15:49
  • $\begingroup$ If the sling shot has 1000J energy when extended, then it can lift the mass 1 m from that point. If the rubber was extended 0.2 m, the object will "fly" 0.8 m. This is further complicated by the question whether "equilibrium position" is the position of the object testing against (and extending) the sling. Is the energy stored at that point included in your 1000 J? $\endgroup$ – Floris Aug 3 '15 at 16:24
  • $\begingroup$ @Floris - Is see what you are saying. Lets suppose for simplicity that the slingshot has 1000 J when fully extended. $\endgroup$ – Matteo Aug 3 '15 at 18:24
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Your $E$ is potential energy in the rubber, which transforms to kinetic $K$. So your starting velocity $v$ will be:

$E=1/2 mv^2$

From conservation of energy:

$0-1/2 mv^2=0-mgh$

$h=v^2/2g$ and $v^2=2E/m$

Confirmed

Interestingly enough the rubber does not obey Hook's law, and you need a lot more work if you want to find out what really happens.https://en.wikipedia.org/wiki/Neo-Hookean_solid

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  • $\begingroup$ So you are saying that my formula is correct? $\endgroup$ – Matteo Jul 30 '15 at 16:26
  • $\begingroup$ I have the impression that your answer (and my question) is not considering the fact that the object is also facing the opposing force determined by gravity. $\endgroup$ – Matteo Jul 30 '15 at 16:32
  • $\begingroup$ The force due to gravity is your potencial energy.$F=-dU/dx$ $\endgroup$ – cianan1705 _c_ Jul 30 '15 at 16:34

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