Im having a hard time to get this question right. A block of mass m is at rest a
ID: 1352910 • Letter: I
Question
Im having a hard time to get this question right.
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(1x/L). That is, the coefficient of friction decreases from 0 at x=0 to zero at x=L.
Question:
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.
Explanation / Answer
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)]