How Does a Variable Coefficient of Kinetic Friction Affect Block Acceleration?

[bestchemist]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

√(2L(F0m−μ0g))+√μ0gL

and it wrong

 

[voko]

What is the problem?

 

[bestchemist]

What is the problem?

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.

 

[Chestermiller]

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.

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.

 

[bestchemist]

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.

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

 

[haruspex]

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

Right so far

I got
√(2L(F0m−μ0g))+√μ0gL

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

 

[bestchemist]

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.

(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?

 

[haruspex]

(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?

Yes, but you can simplify it a little.

 

[bestchemist]

Yes, but you can simplify it a little.

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

 

[Chestermiller]

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

 

[bestchemist]

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

okay.. got it