Simple kinematics with friction

  • Thread starter fishinsea
  • Start date
  • #1
5
0

Homework Statement


A stack of two blocks sits on a frictionless surface; however, between the two blocks is a kinetic coefficient of friction μk. 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


Fnet = 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 - ukg. Solving for t, we get t = √(2x1/(F/m - ukg). Also, vfinal = 2vavg = 2 * (x1/t + x2/t) / 2 = something with uk and g in it.

The smarter method is just to use F = ma and acm = F/2m. The final position of the center of mass is x1+x2/2 and using v2f - v2i = 2ad we get vcm = √F/2m * (x1 + x2) which doesn't involve uk 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.
 

Answers and Replies

  • #2
20
0
For me, the 2nd method is totally fine.

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

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