Question

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 = L across a horizontal surface whose coefficient of kinetic friction is ?_k = ?₀ ( 1 ? x / L ). That is, the coefficient of friction decreases from ?₀ 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 a_x that can be written in terms of v_x and dv_x / dx. 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 v_x and dv_x / dx.

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, ?₀, and appropriate constants.

Answer 1

Apply 2nd law; net force (applied - friction) equal ma;
ma = Fo - f = Fo - umg = Fo - uo(1 - x/L)*mg
a = (Fo/m) - uo(1-x/L)g

a = dv/dt ,
v = dx/dt

We can write -
dv/dt = (dv/dx)(dx/dt) = v(dv/dx)

Acceleration eq is -
v(dv/dx) = (Fo/m) - uo(1-x/L)g

To find v , multiply with dx and integrate the left side from 0 to v and the right side from 0 to x;
vdv = (Fo/m)dx - uog dx + (uog/L)x dx
(1/2)v^2 = (Fo/m)x - uogx + (uog/2L)x^2
v = sqrt[2*((Fo/m)x - uogx + (uog/2L)x^2)]

Velocity of the block when it reaches some position x is given by -
v = sqrt[2*((Fo/m)x - uogx + (uog/2L)x^2)]