# Kinetic friction homework

1. Jul 30, 2013

### bestchemist

1. The problem statement, all variables and given/known data

A block of mass m is at rest at the origin at t=0. It is pushed with constant force F0 from x=0 to x=L across a horizontal surface whose coefficient of kinetic friction is μk=μ0(1−x/L). That is, the coefficient of friction decreases from μ0 at x=0 to zero at x=L.

2. Relevant equations

ax = (dvx/dx)vx
a = (F0 - mg(1-x/L))/m

3. The attempt at a solution
√(2L(F0m−μ0g))+√μ0gL

and it wrong

2. Jul 30, 2013

### voko

What is the problem?

3. Jul 30, 2013

### bestchemist

find an expression for the block's velocity when it reaches position x=L.
Express your answer in terms of the variables L, F0, m, μ0, and appropriate constants.

4. Jul 30, 2013

### Staff: Mentor

Please show your work. It looks like you made some math errors, and you also left out a factor of μ0 in your equation for the acceleration.

5. Jul 30, 2013

### bestchemist

a = (F0 - mg(1-x/L))/m
vdx = (Fo/m)dx - uogdx + (uog/L)xdx
(1/2)v^2 = (Fo/m)x - uogx + (uog/2L)x^2
and solve for v and plug x = L
I got
√(2L(F0m−μ0g))+√μ0gL

6. Jul 30, 2013

### haruspex

Right so far
That's certainly wrong. For others to see where you went wrong you will need to post every step of your working.

7. Jul 30, 2013

### bestchemist

(1/2)v^2 = (Fo/m)x - uogx + (uog/2L)x^2
and solve for v and plug x = L
v^2 = 2 (Fo/m)x - 2 uogx+ 2(uog/2L)x^2
v^2 = 2x ((Fo/m)- uog) + (uog/L)x^2
v^2 = 2L((Fo/m)- uog) + (uogL)

v= √(2L((Fo/m)- uog) + (uogL))

is it right?

8. Jul 30, 2013

### haruspex

Yes, but you can simplify it a little.

9. Jul 30, 2013

### bestchemist

Where can I simplify it? I thought I already simplified it to the most simplest form.
Can you give some hint?

10. Jul 30, 2013

### Staff: Mentor

Check out the terms under the radical (which extends over the entire rhs), and see if you can combine anything.

11. Jul 30, 2013

### bestchemist

okay.. got it

Last edited: Jul 30, 2013