Question

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=Lacross 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.
Part A

We would like to know the velocity of the block when it reaches some position x. Finding this requires an integration. However, acceleration is defined as a derivative with respect to time, which leads to integrals with respect to time, but the force is given as a function of position. To get around this, use the chain rule to find an alternative definition for the acceleration ax that can be written in terms of vx and dvxdx. This is a purely mathematical exercise; it has nothing to do with the forces given in the problem statement.

Express your answer in terms of the variables vx and dvxdx.

I got the answer:

ax =
dvxdxvx

And this was correct, but Im having trouble with Part B:

Now use the result of Part A to 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.

Answer 1

Answer:

A) v = √[(F₀/m - μ₀) 2x + x2 / L] , b) v = √[2 (F₀ / m - μ₀) +1] L

Explanation:

For this exercise we will use Newton's second law

X axis

           Fo - fr = ma

Y Axis  

          N - W = 0

          N = mg

The equation for the force of friction is

         fr = μ N

        .fr = μ mg

Let's replace

     Fo - μ mg = m a

     a = Fo / m - μ g

They ask us the speed of the block, so we must use

     a = dv / dt = dv/dx  dx dt = dv/dx   v

     v dv = a dx

We replace

     v dv = (F₀ / m - μ₀ (1- x/L)) dx

We integrate

     ∫v dv = F₀ / m ∫ dx - μ₀ ∫ dx + 1 / L ∫ x dx

    ½ v² = (F₀ / m - μ₀) x + 1 / L x² / 2

We evaluate from the lower limit, for x = 0 the speed is zero v = 0, to the upper limit x, v

     ½ v² = (F₀ / m - μ₀) x + 1 / 2L x²

     v = √[(F₀ / m - μ₀) 2x + x2 / L]

Part B

The velocity for x = L

   v = √(F₀ / m - μ₀) 2L + L2 / L

   v = √[2 (F₀ / m - μ₀) +1] L