1. A 5 kg block rests on a horizontal table, attached to a 4 kg block by a light string as shown in the figure. The acceleration of gravity is 9.81 m/s 2 . The coefficient of kinetic friction between the block and the table is 0.3, find the time it takes for the 4 kg mass to fall 7 m to the floor if the system starts from rest. Answer in units of s. 2. Relevant equations 5 kg block = B 4 kg block = A u = 0.3 g=9.81 m/s^2 Ff = Force of Friction = u*mass*acceleration F = Force = mass*acceleration 3. The attempt at a solution I started by finding the acceleration of the 5kg block using the following equation: FB = mB * a = FA - Ff Solving for FA: FA = (4 kg)(9.81 m/s^2) = 39.24 N solving for Ff: Ff = (0.3)(5 kg)(9.81 m/s^2) = 14.715 N solving for acceleration (a): (5 kg) * a = (39.24 N) - (14.175 N) (5 kg) * a = 25.065 N a = (26.065 N) / (5 kg) = 4.905 m/s^2 solving for time (t) to reach 7 meters: (a*t^2)/2 = 7 m a*t^2 = 14 m t^2 = (14 m) / (a) t = sqrt [(14 m) / (a)] ≈ 1.689 seconds The answer comes back as incorrect when I submit it. Any help would be appreciated.