pls !
a string is tied to a 4.4 kg block and 120g hanging bucket. students add 20g washers one at a time to the bucket. the student are unaware that the
coefficient of static friction for the block on the table is 0.42.
a) what is the maximum force of static friction for the block?
18.11n
b) how many washers can the students add to the bucket without moving the block?
86
c) the coefficient on kinetic friction is 0.34. calculate the acceleration of the block when the final washer is added to the bucket and the objects start to
the diagram is the block on the table that is sitting horizontally on a table with a pulley holding the bucket vertically with one string. the block and the bucket are tied with one string.
i jus didn't get c)

I'm slightly dubious that you solved A). and B). on your own but you're having trouble with C).
In order to save myself a lot of work, I'll take your word for it.  The answers shown there for A). and B). are correct, so I'll go directly to C).

The block breaks free of static friction and starts moving when there are 86 washers in the bucket.

Total mass of the bucket and washers is (0.120 kg) + 86(0.02 kg) = 1.84 kg

Total weight of the bucket & washers = (M) (G) = (1.84 x 9.8) = 18.04 N
That's the tension in the string, sliding the block along the table towards the edge.

How much force of friction is pulling the block back, not letting it move ?
It's (weight of the block) x (coefficient of kinetic friction)

       (M) (G) x (0.34) = (4.4) x (9.8) x (0.34) = 14.66 N

The block has 18.04 N pulling it forward, and 14.66 N holding it back.
What's the net force on the block ?

   18.04 - 14.66 = 3.38 N     forward, toward the pulley.

Take the magic formula that always connects acceleration with net force:

       F = M A

You know 'F' and 'M' and you need to find 'A', so Divide each side by (M):   

     A = F / M         and there you have it for the block.

     A= F / M = (3.38 N) / (4.4 kg) = 0.768 meter / second-squared

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