What Spring Constant is Needed for a Spring-Launched Roller Coaster?

[cyclemun]

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 12--high hill, then descends 20 to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.3 and that a loaded car will have a maximum mass of 410 . For safety reasons, the spring constant should be 15 larger than the minimum needed for the car to just make it over the top.


What spring constant should you specify?
What is the maximum speed of a 370 car if the spring is compressed the full amount?

I don't even know where to begin.

 

[LowlyPion]

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 12--high hill, then descends 20 to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.3 and that a loaded car will have a maximum mass of 410 . For safety reasons, the spring constant should be 15 larger than the minimum needed for the car to just make it over the top.

What spring constant should you specify?
What is the maximum speed of a 370 car if the spring is compressed the full amount?

I don't even know where to begin.

Welcome to PF.

I'd start with wanting to know what the units were.

 

[cyclemun]

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 12m high hill, then descends 20m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.3m and that a loaded car will have a maximum mass of 410kg . For safety reasons, the spring constant should be 15% larger than the minimum needed for the car to just make it over the top.What spring constant should you specify?
What is the maximum speed of a 370kg car if the spring is compressed the full amount?

 

[cyclemun]

based on my logic, it shouldn't matter what happens after the cart makes it over the hill so i would just use the formula .5kx^2 = mgh

 

[LowlyPion]

based on my logic, it shouldn't matter what happens after the cart makes it over the hill so i would just use the formula .5kx^2 = mgh

It would equal 1.15 * m*g*h wouldn't it?

As you note then h you are looking at is 12 m.

Then the max speed would be the excess KE at the top plus its potential turned to KE at the bottom of the 20 m.

 

[cyclemun]

how would i find the excess KE at the top? is that just the original mgh of the 410 kg cart minus the mgh of the 370 kg cart?

 

[cyclemun]

ok i got it the max speed is (410)(9.8)(12) - (370)(9.8)(12) + (370)(9.8)(20) = .5 (370) v^2 and i get a max speed of 21 m/s

 

[LowlyPion]

how would i find the excess KE at the top? is that just the original mgh of the 410 kg cart minus the mgh of the 370 kg cart?

Actually your spring constant should incorporate all the necessary energy for your calculation. If you calculated the k at 115% for the 410 cart then that spring constant less the energy to push the 370 kg cart up 12 m is your excess. Then you add back the 20 m of gravity to give you the KE. Your method has only calculated getting the 410 kg cart to the top and not the additional 15% you were asked to put into the k of the spring.

 

[LowlyPion]

ok i got it the max speed is (410)(9.8)(12) - (370)(9.8)(12) + (370)(9.8)(20) = .5 (370) v^2 and i get a max speed of 21 m/s

Alternatively you can increase the (410)(9.8)(12) term by the factor of 1.15 as this is what you should have used to calculate the k of the spring.