Simple kinematics with friction

[fishinsea]

Homework Statement

A stack of two blocks sits on a frictionless surface; however, between the two blocks is a kinetic coefficient of friction μ_k. External force F is applied to the top block. During the time the force is applied, the top block is displaced by x1, and the bottom block is displaced by x2. Assume enough force is applied that x1 > x2. What is the final velocity of the center of mass of the system in terms of the values above and g?

Homework Equations

F_net = ma and the standard kinematics equations

The Attempt at a Solution

Since force is constant, acceleration is constant and x1 = 1/2at² where a = F/m - u_kg. Solving for t, we get t = √(2x1/(F/m - u_kg). Also, v_final = 2v_avg = 2 * (x1/t + x2/t) / 2 = something with u_k and g in it.

The smarter method is just to use F = ma and a_cm = F/2m. The final position of the center of mass is x1+x2/2 and using v²_f - v²_i = 2ad we get v_cm = √F/2m * (x1 + x2) which doesn't involve u_k or g at all.

I'm wondering whether there's a the top method is wrong or if there's a simplification step that would link the two answers.

 

[cheah10]

For me, the 2nd method is totally fine.

For the 1st method, can you explain whether it is a=(F/m)-u_kg or it is a=F/(m-u_kg)?