Question

the kinetic coefficient of friction for a 12 kg mass on a table is 0.25. a string from this mass goes across a pulley and is attached to an 8.0 kg mass hanging over the table. if the system is moving so that the hanging mass is moving downward, find the magnitude of the acceleration of each mass and the tension in the string

Answer 1

The friction acts on the block on the table in backward direction and tension force in the forward direction

On the block hanging

weight acts downwards direction and tension in the string in the upward direction

So

mg - T = ma ............(hanging block)

8*9.8 - T = 8a............(1)

T - uN = Ma...................(block on table)

T - 0.25*12*9.8 = 12a...............(2)

Adding 1 and 2

78.4 - 29.4 = 20a

a = 49/20 = 2.45 m/s²

Substituting in 1

78.4 - 8*2.45 = T

T = 58.8 N