How Do You Calculate Kinetic Energy and Velocity on a Roller Coaster?

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To calculate kinetic energy (KE) and velocity on a roller coaster, start by determining the potential energy (PE) at the top of each hill using the formula PE = mgh, where m is mass, g is gravity, and h is height. For the first hill at 40m, the PE is 196,000J, which equals the KE at the bottom of that hill, assuming no energy losses. Use the equation KE = 1/2mv² to find velocity, rearranging it to solve for v after calculating KE. The conservation of mechanical energy principle states that the total energy remains constant, allowing for the calculation of KE and velocity at various points. Ensure to create a diagram to visualize the energy transitions throughout the coaster's path.
bls555
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okay, there is a roller coaster with 2 hills.. this is what I know:
1st hill height: 40m
2nd hill height: 25m
mass of coaster: 500kg
I was asked to find out the PE at top of hill #1, I did that and got 196,000J
Now I have to findthe KE at the bottom of hill #1, as well as finding the velocity at various spots, how do I do this?
Thanks a LOT for the help in advance.
 
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Make a list of the values for the PE and KE of the system at each point you want to evaluate. (make sure you set up an appropriate diagram,, from your first answer I assume your x-axis lies along the ground)

Then make use of the equation for conservation of mechanical energy

KE_intial + PE_initial = KE_final + PE_final

do you know the equations for KE and PE?
 
Last edited:
PE = mgh
KE = 1/2mv²

I don't get how I'm supposed to find KE without knowing velocity, or without knowing T to find velocity (V = d/t).
EDIT: Yeah, each hill starts and ends at ground level, sorry for not including that.
 
Last edited:
At the bottom of hill one the KE will equal the PE at the top of hill one. One you get that you can calculate the velocity by solving Ek=1/2mv^2 using the mass of the coaster and KE that you just found.

This is of course assuming no losses
 
Thank you!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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