A 5.7- kg concrete block rests on a level table. A 4.0- kg mass is attached to the block by a string passing over a light, frictionless pulley. If the acceleration of the block is measured to be 1.0 m/s2, what is the coefficient of friction between the block and the table?

Answer 1

The net force pulling the 5.7 kg block is
Fnet = M1*a = 5.7*1.0 = 5.7 N

That equals the string tension M2*g minus the friction force M1*g*U
5.7 = M2*g - M1*g*U

M1 = 5.7 kg
M2 = 4.0 kg
Uk = kinetic friction coefficient
g = 9.8 m/s^2

Solve for U

Answer 2

Drwls has it right, but make sure that your Fnet = (M1+M2)*a