A 28.0 kg block is connected to an empty 1.00 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.45 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system begins to move. a) Calculate the mass of sand added to the bucket b) Calculate the acceleration for the system. a) 28.0 kg block weighs (28.0 kg)(9.80 m/s2) = 274.4 N Ffr(max) = µsFN Ffr(max) = (.450)(274.4 N) = 123.48 N F = ma 123.48 N = m(9.80 N/kg)=12.6 kg 12.6 kg – 1.0 kg=11.6 kg of sand b) Force of Kinetic Friction between Block and Table Ffr = µkFN Ffr = (.320)(274.4 N) = 87.808 N Force of Bucket F=ma 123.48 - T= 12.6a Force of Block F = ma T - 87.808 N= 28a Substituting First Expression into Second Expression and Solve for T: 123.48 - T = 12.6a, T = 123.48 - 12.6a T - 87.808 N = 28a (123.48 - 12.6a) - 87.808 N = 28a 35.672 = 40.6a a = 0.879 m/s2 Does this all look correct?