Help needed in creating a physics equation

SnowDancer

Hi,
I’m not sure this is the right place to post this but I am hoping someone can help me come up with a mathematical equation.

This image (attached), I hope, is self-explanatory.

What I need to know is, if the container is filled with 1,500 cubic metres of water in 5 seconds and the rope is then cut. How much energy is imparted (on the rock) and is there a simple formula so that I can do this for different values?

Thanks in advance
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im invisible

1500 cubic metres of water? in 5 seconds??

thats 300 tonnes a second....
(i dont think the time matters at all, but still, somethings going to break)

you need to know the elasticity of the rope (in the slingshot) first, unless i'm reading it wrong, and a very big container, to fire a relatively small rock...

kirving

Maybe 1.5 cubic meters of water, or 1500 litres?

Anyway, it doesn't matter how quickly the tank is filled, the force acting on the elastic is the same. 1,500 [m^3] * 1,000[kg/m^3] = 1,500,000 [kg]

Force = Mass x Acceleration
F=ma


Multiply by Gravity, g, to get Force in Newtons. 1,500,000 [kg] * 9.81[m/s^2] = 14,715,000[N] This is the force created by the weight of the water tank.

Equal and opposite reactions etc, etc, so: Force down due to weight of water (I'm assuming rock is held in place) = Tensile force in rubber band/bending force on slingshot shaft.

F=ma
14,715,000[N] = 500[kg] * Acceleration

Acceleration on rock when rope is cut = 29430 [M/s^2], which is mad.

Also, F=kx (for the spring effect of the rubber bands), but I'm not sure that this matters for this question.

See, the unit for energy is the Joule, but as far as I remember, 1 Joule = 1 Newton x 1 Meter, so I'm not sure how this comes into play for the question.

SnowDancer

Thank you both :-D

citrus burst

Are there any lengths given?

Any way the potential energy from the rope is given by [Latex]\frac{1}{2}kx^2[/Latex]

And the potential from gravity is [Latex]mgx[/Latex]

Trick is to work out the work done in bringing the rock to where ever. That should be equal to the potential energy stored in the rock/rope.

FISMA

Kevin Irving wrote: »

Anyway, it doesn't matter how quickly the tank is filled, the force acting on the elastic is the same.

That depends on when we are looking at the stretch of the elastic.

Consider jumping on a scale. Does the scale read more than your weight? Clearly it does.

The scale has to do two things: support the weight of a mass and change the momentum of that mass.

The greater the volume flow rate, the more the elastic will stretch. Once, however, the water is finished filling the container, it will of course reach an equilibrium position, which is what I think Kevin Irving was referring to.

yubabill1

Trick is to work out the work done in bringing the rock to where ever. That should be equal to the potential energy stored in the rock/rope.

Don't see how that could be done, without making assumptions.

It may be OK to make assumptions, but that seems a little advanced given the nature of the problem posed. I would much rather have more info.

There doesn't seem to be a way to calculate what the OP wants to know, with the info provided IMHO.