A block of mass M is at rest at the origin at t=0. It is pushed with a constant

Question

A block of mass M is at rest at the origin at t=0.  It is pushed with a 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.

A.) Use wht yu've leaned in calculat to prove that Ax = Vx DVx

                                                                      Dx

Explanation / Answer

So let's identify all of the forces being applied on the block. You have a Force Normal that is equal to it's weight since it is not accelerating in the vertical. FN = W = mg You have a constant force pushing the block identified as F0 and a Friction Force resisting that movement. Friction Force = (FN)*(COEFFICIENT OF FRICTION) Friction Force = mg(1-x/L) Therefore the sum of all forces in the horizontal are F0 - mg(1-x/L) = ma Now solve for acceleration and we get: (F0 - mg(1-x/L))/m = a Now you have an acceleration function that is dependent on the value of x. --> a(x) In order to find the velocity function, we simply integrate with respect to x, where F0, m, g, and L are constants.

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