What is the force required to bring a roller coaster to a stop at point E?

[tascja]

Force and Energy Question!

Homework Statement

the force that must be applied to bring it to a stop at point E
(this is only one part of a multi-part question - the rest of the information: Consider the frictionless rollar coaster. The height at A (the start of the rollar coaster) is 95m, it is downward and then upward to C, at 65m high, then down again and then up again to 25m at D until a downward slope to a stop at ground level, point E.
[sorry for the written explanation, but a picture needs approval - takes too long lol]

Homework Equations

Et = Ek + Ep
W = Fd
W = Ek

The Attempt at a Solution

Et = Ek + Ep
Ek = Et - Ep
Ek = Et - 0
Ek = 11172000 J {calculated in previous question}

W = Fd
F = W/d
F = Ek/d
F = 11172000 J/ ?

** so my dilemma is that it does not give me the distance in the question. Is there any other equations or methods that I could use instead to solve for the force??
My thanks for your help in advance!

 

[aniketp]

so my dilemma is that it does not give me the distance in the question. Is there any other equations or methods that I could use instead to solve for the force??

Try using the work kinetic energy thm which states that
\DeltaW =\DeltaK.E

 

[tascja]

but then how will i find F (force)?? because Ek = 1/2mv^2

 

[cryptoguy]

How did you figure out Ek if you are not given the mass of the object?

 

[tascja]

i am given mass, sorry i forgot to include it, its 12000 kg
so:

Ek = Et - Ep
Ek = 11172000 - mgh
Ek = 11172000 - (12000)(9.8)(65)
Ek = 3528000 J

 

[cryptoguy]

ok I see, so you found the potential energy on top of the 95 m hill and then found the kinetic energy on the 65 m hill. You need to do this again for the 25 m hill, and finally find the final speed of the cart.

hmm, you're not given the stopping distance? Nor the time it takes for the cart to stop? Is there any other information at all?

 

[ace123]

Maybe the stopping distance can be determined from your diagram. Try to upload it or take a closer look yourself.

 

[tascja]

no distance is given and no stopping time but ill upload a picture of the track

 

[tascja]

http://www.geocities.com/tascja99/physics.html?1215530828005"

 

[tascja]

What if i were to just use F=mg, because it is a downward slope plus there is no friction acting on it to consider?

 

[cryptoguy]

hmm afraid not because F = mg is the equation for free-fall but the object here is certainly not in free fall and is therefore not accelerating at -9.8 m/s^2