Question

A 29.5 kg block is connected to an empty 1.00 kg bucket by a cord running over a frictionless pulley (Fig. 4-57). The coefficient of static friction between the table and the block is0.420 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 just begins to move.

The mass of the sand added to the bucket is 11.39 kg

Calculate the acceleration of the system (downward) in m/s2 (use the coefficent of kinetic friction)

Answer 1

let tension in the cord is T and acceleration of the system is a m/s^2.

then for the bucket, writing force balance equation:

(11.39+1)*9.8-T=(11.39+1)*a

==>121.422-T=12.39*a....(1)

for the block on the table, friction will be kinetic friction and it will act to the left in order to oppose the motion.

normal force on the block=weight of the block=29.5*9.8=289.1 N

then kinetic friction force=kinetic friction coefficient*normal force=0.32*289.1=92.512 N

writing force balance equation for the block:

T-92.512=29.5*a...(2)

adding equation 1 and 2:

121.422-92.512=(12.39+29.5)*a

==>a=0.69 m/s^2

hence acceleration of the system is 0.69 m/s^2, downward.