What is the roller coaster's final speed at the bottom?

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irrrjntlp
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Homework Statement



A roller coaster is lifted up 50m above the ground to the top of the first hill and then glides down around the track at the bottom. If it had a velocity of 3.0 m/s at the top of the lift and loses 10% of its total energy to friction as it glides down, what is the roller coaster's final speed at the bottom. Sorry, I can't remember the mass (in the question, i didn't actually forget it).

Thanks for any help!

Homework Equations



v2^2 = v1^2 + 2g (y1-y2) <- not sure if this is correct..

The Attempt at a Solution



v2^2 = 3^2 + 2(9.8)(50-0)
v2^2 = 9 + 980
v2^2 = 989
V2 = 31.4 m/s

I'm not sure how to apply the friction acting against without a mass...

9.8 - (9.8 x .1)
8.82 <- Maybe use this as acceleration instead..
 
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It would be beneficial to approach this problem from an energy standpoint. How much energy was added to the system by raising the car up to 50 m? What subsequently happened to that energy? Where did it go?
 
irrrjntlp said:

Homework Equations



v2^2 = v1^2 + 2g (y1-y2) <- not sure if this is correct..

Not correct.

irrrjntlp said:
I'm not sure how to apply the friction acting against without a mass...

Just assume some mass m. Use the energy approach as advised in the previous post.

Use the fact that initial mechanical energy must be equal to final mechanical energy minus loss due to friction.