Two blocks are arranged at the ends of a massless string. A 4.85 block fixed to one end is kept on a table and on the other end a 3.2 kg block can fall freely. The system starts from rest. When the 3.2 kg mass has fallen through 0.399 m, its downward speed is 1.28 m/s.
Use the expression relating initial velocity u, final velocity v, acceleration a and the distance traveled s, v^2 - u^2 = 2*a*s
Here u = 0, v = 1.28 and s = 0.399
a = 1.28^2/(2*0.399) = 2.05 m/s^2
The block of mass 3.2 kg that is falling has a force of gravity equal to 3.2*9.8 N acting on it. This is being countered by the frictional force between the other bock and the table.
3.2*9.8 - F = 3.2*2.05
=> F = 24.78 N
The frictional force between the 4.85 kg mass and the table is 24.78 N