# Particle sliding across a surface with curved ends

1. Mar 15, 2009

### rmunoz

1. The problem statement, all variables and given/known data
A particle can slide along a track with elevated ends and a flat central part, as shown below. The flat part has length L = 40 cm. The curved portions of the track are frictionless, but for the flat part the coefficient of kinetic friction is µk = 0.28. The particle is released from rest at point A, which is a height h = L/2. How far from the left edge of the flat does the particle finally stop? http://www.webassign.net/halliday8e/pc/halliday8019c08/halliday8019c08-fig-0057.htm [Broken]

2. Relevant equations
Ugrav=mgh
K= 1/2mv^2
a(centripital)= v^2/r
Fric(kinetic)= $$\mu$$mgsin90=> $$\mu$$mg

Sorry to say, but the reason i'm having trouble with this problem is frankly that I have no idea how to start! The goal for me is to be able to use the law of conservation of mechanical energy to describe the particle's motion. At first glance, I feel like the equation for the kinetic energy of the particle, right after the first fall should take into account the velocity calculated by centripital acceleration. I can tell initially that the kinetic energy will be 0 since its not moving and that the Ugrav will describe the total mechanical energy within the system. Is this correct, or even relevant at all? Also, i have no idea how to describe the loss of kinetic energy as an increase in the amount of "heat" energy within the system due to the friction-surface. Any tips on how to get the ball rolling on this problem?

Sorry if this is at all unclear, this is my first post and as you can probably tell, i'm in dire need of some physics help, but more than willing to put in the effort with guidance!

Last edited by a moderator: May 4, 2017
2. Mar 15, 2009

### tiny-tim

Welcome to PF!

Hi rmunoz ! Welcome to PF!

As you say, this is an energy problem …

use conservation of energy on the curved part (the centripetal acceleration is irrelevant), and the work-energy theorem ( work done by friction = energy lost) on the flat part

3. Mar 15, 2009

### rmunoz

Wow, got it right on the first try! that was a lot easier than i thought it was gonna be, thank you very much tiny-tim for the help... i think i like this website!