A mass m1 on a horizontal shelf is attached by a thin string that passes over a frictionless pulley to a 3.0 kg mass (m2) that hangs over the side of the shelf 1.3 m above the ground. The system is released from rest at t = 0 and the 3.0 kg mass strikes the ground at t = 0.78 s. The system is now placed in its initial position and a 1.2 kg mass is placed on top of the block of mass m1. Released from rest, the 3.0 kg mass now strikes the ground 1.3 seconds later. (a) Determine the mass m1. kg (b) Determine the coefficient of kinetic friction between m1 and the shelf. What I done: d = ½at² In this case: 1.3m = ½(a_before)(0.78s)² and: 1.3m = ½(a_after)(1.3s)² Ab = 4.3m/s^2 Aa = 1.5m/s^2 In the "before" case: 1)F_net1_before = T_before - (μk)(m1)(g) = (m1)(a_before) and: 2)F_net2_before = (m2)(g) - T_before = (m2)(a_before) In the "after" case: 3)F_net1_after = T_after - (μk)(m1+1.2kg)(g) = (m1+1.2kg)(a_after) and 4)F_net2_after = (m2)(g) - T_after = (m2)(a_after) I combined equations 1 and 2 to eliminate Tb I combined equations 3 and 4 to eliminate Ta I plugged in the numbers and got up to 16.53 - (Uk)(M1)(9.81) = (M1)(4.3) and 20.43 -(Uk)(M1+1.2)(9.81) = (M1+1.2)(1.5) First equation, solved for Uk: Uk = (16.53 - (M1)(4.3)) / ((M1)(9.81)) Second equation, solved for Uk: Uk = (20.43 - (M1+1.2)(1.5)) / (M1+1.2)(9.81) (16.53 - (M1)(4.3)) / ((M1)(9.81)) = (20.43 - (M1+1.2)(1.5)) / (M1+1.2)(9.81) I did that and ended up with quadratic equations and I plug that in the formula and ended up with a value for M1. When I plug in M1 to the 2 Uk equations, they did not match. What I did wrong? Please help, I spent hours on this, please help. Thanks!!! I believe is more of algebraic mistake but Ive been going through this so much, if anyone can point out where I went wrong.