Friction Problem -- String, Block, Bucket & Washers

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SUMMARY

The discussion centers on a physics problem involving static and kinetic friction with a 4.4 kg block and a 120g bucket. The maximum force of static friction is calculated to be 18.11N, allowing for the addition of 86 washers before the block moves. The coefficient of kinetic friction is given as 0.34, leading to a calculated acceleration of 0.52 m/s² when the final washer is added. There is a discrepancy noted between the calculated acceleration and the book's answer of 0.75 m/s², suggesting a potential typo in the textbook.

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Nithya115
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Homework Statement


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[/B] 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?
Answer: 18.11N
B) how many washers can the students add to the bucket without moving the block?
Answer: 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 move...

The diagram is the block on table with the pulley on the edge of the table holding the bucket vertically. They are tied with one string.
How do get c)? I got 0.56m/s^2 but the answer in the book is 0.75m/s^2

Homework Equations


a= m2g -Fs/ m1 + m2

The Attempt at a Solution


a= m2g -Fs/ m1 + m2
a= (1.86 x 9.8) - 15/ (4.4 + 1.86)
a= 0.52m/s^2
I got Fs by...
Fs = (0.34)(Fn=4.4x9.8)
= 1.86
I got m2 by...
1. calculating the mass of the washers
20g x 87 washers = 1740g
2. adding the bucket and washers
120g + 1740g = 1860g
3. converted to kg
1860g/1000g=1.86 kg
 
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You got 0.56 but in 3) you use 15 instead of 14.66 and mention 0.52 m/s2. Confusing.

Nevertheless, I get 0.57 and suspect a typo in the book answer
 

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