# Friction/work/energy question

1. Oct 17, 2004

### StonieJ

Three identical blocks move in three different settings. Block A moves on a horizontal surface, block B moves up an inclined plane, and block C moves down an inclined plane. They all start with the same speed and continue until brought to rest by friction. Rank which block dissipates more mechanical energy by friction, least to greatest.

I am not very good with these "conceptual" questions, so I'm having a hard time even finding out where to start. I've been trying to calculate the forces of friction for each block (e.g. fA = u * mg, fB = u * mg * cos(T), etc) but it hasn't helped much. So, I was hoping somebody could at least give me a push in the right direction.

2. Oct 17, 2004

### Galileo

The frictional force is proportional to the normal force on the block.
In which situation is the normal force greatest?

3. Oct 17, 2004

### StonieJ

Well, your answer is encouraging because that's what I was doing originally, i.e. calculating the forces of friction on each block.

Block A
fA = u * mg
For block A, the normal is simply the mass times acceleration due to gravity.

Block B
fB = u * mg * cos(T)

Block C
fC = u * mg * cos(T)

For blocks B and C, the normal force is only a fraction of the mass times acceleration due to gravity.

So, based on that, I would say that block A dissipates energy the fastest and B and C dissipate energy equally. It's just that, intuitively, I have a hard time seeing this. Intuitively, I would be inclined to say that the block going uphill dissipates energy the fastest, followed by the block on the flat surface, followed by the block going down the plane.

4. Oct 17, 2004

### Galileo

The block going uphill will probably come to a halt faster, but that doesn't mean it dissipates energy faster. It gains potential energy in return for kinetic energy. (it stops faster, but not due to friction).
You're asked for the dissipation in energy due to friction, so all you need is the magnitude of the frictional force.
I agree it's not very intuitive, but many things in classical mechanics aren't even though you'd expect otherwise.

5. Oct 17, 2004

### wisky40

I think you can use kinetic energy and potential energy.
For block A the kinetic energy (1/2)mv^2= energy dissipated on friction(onA).
For block B the kinetic energy (1/2)mv^2= energy dissipated on friction(on B) +mgh' => (1/2)mv^2-mgh'= energy dissipated on friction(onB).
Finally, for block C (1/2)mv^2+mgh"= energy disspated on friction(onC).
kinetic energy is the same for the three cases because the have the same velocity and mgh'>0 and also mgh">0=>
energy dissipated on B< energy dissipated on A< energy dissipated on C.