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

AI Thread Summary
To determine the roller coaster's final speed at the bottom after descending from a height of 50 meters, it starts with a velocity of 3.0 m/s and loses 10% of its energy to friction. The initial mechanical energy can be calculated using gravitational potential energy and kinetic energy. The correct approach involves equating the initial mechanical energy to the final mechanical energy minus the energy lost to friction. By applying these principles, the final speed can be accurately calculated, taking into account the energy lost due to friction.
irrrjntlp
Messages
1
Reaction score
0

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..
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top