Problem 6.72 A block of mass m is at rest at the origin at t = 0. It is pushed w
ID: 1318709 • Letter: P
Question
Problem 6.72 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 muk = mu0(1-x/L). That is, the coefficient of friction decreases from mu0 at x = 0 to zero at x= L alternative definition for the acceleration at that can be written in terms of vx and 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 dvx/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, mu0, and appropriate constants.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 , but you want to express this in terms of x ,rather then t, so use the chain rule;
dv/dt = (dv/dx)(dx/dt) = v(dv/dx)
so your acceleration eq is;
v(dv/dx) = (Fo/m) - uo(1-x/L)g
to find v , multiply thru the dx and integrate the left side from 0 to v and the right side from 0 to x;
vdv = (Fo/m)dx - uogdx + (uog/L)xdx
Integrate
(1/2)v^2 = (Fo/m)x - uogx + (uog/2L)x^2
You can do the rest. Just solve for v in terms of x , then evaluate it at x=L if you want the velocity at that location.
(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))