Three roller-coaster carts with a combined mass of 1250 kg have a speed of 15.0 m/s nearing position D.
As the carts approach position D they are stopped by a large hydraulic shock absorber that acts as a spring and compresses a distance of 2.00 m. Determine the spring constant of the shock absorber. Show all work. PICTURE IS BELOW IN COMMENTS

Question

Three roller-coaster carts with a combined mass of 1250 kg have a speed of 15.0 m/s nearing position D.  
As the carts approach position D they are stopped by a large hydraulic shock absorber that acts as a spring and compresses a distance of 2.00 m.  Determine the spring constant of the shock absorber.  Show all work. PICTURE IS BELOW IN COMMENTS

anonymous · Answer

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festinger · Answer

This is a question hinges about the conservation of energy. We first ask how much kinetic energy the car has before collision with the spring. Since the cart stops, the final car does not have any kinetic energy.

There are many ways to express the conservation of energy, and textbooks/teachers aren't terribly clear as to which to use, but I usually start with:

$$KE_{final}+PE_{final}=KE_{intial}+PE_{initial}$$

note that all energy terms that are 'initial' are on one side, and 'final' are on the other. The potential energy term could be stored as elastic(EPE) or gravitational, so be careful that you include all the terms. Around point D, the track is a straight line, therefore we can ignore the Gravitational potential energy(GPE) term: we can put it in, but it drops out since the value does not change:

$$KE_{final}+EPE_{final}+GPE_{final}=KE_{intial}+EPE_{initial}+GPE_{initial}$$
since $GPE_{final}=GPE_{initial}$,
$$KE_{final}+EPE_{final}+GPE_{inital}=KE_{intial}+EPE_{initial}+GPE_{initial}$$
$$KE_{final}+EPE_{final}=KE_{intial}+EPE_{initial}$$
which was the claim i made. These steps are usually ignored, but sometimes helps with understanding the problem.

So, what you need to do is to calculate the $KE_{initial}$ and $KE_{final}$, with the formula
$$KE=\frac{1}{2}mv^{2}$$
In partitcular, $KE_{final}=0$ since $v_{final}=0$ and $EPE_{initial}=0$, sincce the spring is unstretched.

Substituting these and $EPE=\frac{1}{2}kx^{2}$, where $x=2m$ should solve for k.

anonymous · Answer

im sorry im still confused @festinger

anonymous · Answer

which formula do i use"?

festinger · Answer

Ultimately, the equation should reduce to:
$$\frac{1}{2}\times1250\times15^{2}=\frac{1}{2}\times k\times 2^{2}$$

anonymous · Answer

thanks!