Rollercoaster& velocity problem

[chazgurl4life]

Suppose the roller coaster in Fig. 6-41 (h1 = 38 m, h2 = 12 m, h3 = 20) passes point 1 with a speed of 1.40 m/s. If the average force of friction is equal to one third of its weight, with what speed will it reach point 2? The distance traveled is 35.0 m.
m/s


Please view atachment

I don't even know where to start with this problem

 

[Hootenanny]

I'm afraid you problem isn't going to get answered ASAP, it takes a while for attachments to be approved. Perhaps if you described your problem?

 

[PPonte]

Or maybe you could host it in http://www.imageshack.us/ ...

 

[chazgurl4life]

ok so there is a rollercoaster on the ramp and there are 3 diffeternt positions allocated (hi= 38m, H2=12m, H3=20m). Now according to the problem there is a friction force on the rollercoaster which amount to 1/3 of thr rollercoaster's weight. But they don't give the weight of the rollercoaster. Since there is a long distance between hi and h2 they want to know what's the velocity at point 2 ( which is btween H1 &2). My porblem is that i don't even know what equations i should be using because my prof always gave us similar problems in which mass was given. is so i would use :

mgd=mgd2 + friction(distance)

 

[Hootenanny]

mgd=mgd2 + friction(distance)

That is the correct equation to use. Think about how else you could write the frictional force. What other equations do you know for friction?

 

[chazgurl4life]

force of friction = coefficient of friction X normal force
coefficient of friction mass(9.8)
Force of Friction= coefficient of friction x mgcostheta
with the other equation : friction =mgd/mgd2
-----------
distance

 

[Hootenanny]

F_{friction} = \mu R = \mu mg

Substitute F_{friction} = \mu mg into your equation and you will find that the masses cancel...

 

[chazgurl4life]

umg=mgd/mgd2
u=d/d2

 

[PPonte]

chazgurl4life,

I think you did a mistake:

umg=mgd/mgd2
um = d/d2

Only the masses in bold cancel and g.

 

[chazgurl4life]

oh ok thanx