1. The problem statement, all variables and given/known data The 3kg collar is initially at rest and is acted upon a force Q which varies by the the graph shown. Knowing that the coefficient of kinetic friction is 0.25, determine the velocity of the collar at t = 1 s and t = 2s. 2. Relevant equations FΔt = mΔv Ffriction = .25 x N 3. The attempt at a solution I know that the answer for b = 3.43 m/s, but I'm trying to think of how I would get there. I thought about the impulse equation so I tried plugging it in. Since at t = 2, F = 5, and Δt = 2, I said 10 = 3(Vfinal - VInitial where V initial would be 0. I get that the velocity at t = 2 is 3.33, which is close, but no cigar. I then thought about the the force of friction that would be acting in the opposite direction of Q, so I should subtract the two forces. I found that the force of friction is 7.3575 N. Once i subtract the two, I get a negative value, which I don't think is right. Any help would be much appreciated!