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This was a problem on an exam I had the other day and I want to know if my attempt was correct.

Please look at the attached image. You have a system of two ideal pulleys set up on a table. Block A is hanging from a string on the left side of the table which passes by a pulley connecting it to block B. Block B is resting on the table, and the kinetic coefficient of friction between the surface of the table and the block is 0.557. Block C is hanging from a string on the right side of the table which passes by a pulley and is attached to Block B. Block C is four times as massive as block A, and block B is three times as massive as block A. When the system is released from rest, block C accelerates downward until it reaches the floor, a distance of 105cm.

Find: The acceleration of block C and the time it takes to reach the ground.

F=ma

F

y = y

N = mg

I started by trying to find the net force acting on block B in order to find the acceleration of the system. In this case,

F

so

F

simplify

F

because m

F

so the acceleration of block B would be g(1-[tex]\mu[/tex]), and plugging in the values I got the answer -4.34m/s

Next I needed to find the time it took block C to hit the ground.

y = y

so

y = 1/2 at

t = sqrt(2y/a)

t = sqrt(2*-1.05m/-4.34m/s

t = 0.695s rounded to three figures

Did I do this problem correctly? And if not, can you please show me the correct way of doing a problem like this? Thanks.

## Homework Statement

Please look at the attached image. You have a system of two ideal pulleys set up on a table. Block A is hanging from a string on the left side of the table which passes by a pulley connecting it to block B. Block B is resting on the table, and the kinetic coefficient of friction between the surface of the table and the block is 0.557. Block C is hanging from a string on the right side of the table which passes by a pulley and is attached to Block B. Block C is four times as massive as block A, and block B is three times as massive as block A. When the system is released from rest, block C accelerates downward until it reaches the floor, a distance of 105cm.

Find: The acceleration of block C and the time it takes to reach the ground.

## Homework Equations

F=ma

F

_{F}=[tex]\mu[/tex]_{k}Ny = y

_{0}+ v_{0}_{y}t + 1/2 at^{2}N = mg

## The Attempt at a Solution

I started by trying to find the net force acting on block B in order to find the acceleration of the system. In this case,

F

_{net}= F_{C}- (F_{A}+ F_{F})so

F

_{net}= (4m_{A}g) - 1m_{A}g - ([tex]\mu[/tex]3m_{A}g)simplify

F

_{net}= 3m_{A}g(1-[tex]\mu[/tex])because m

_{B}=3m_{A}you can rephrase the above equationF

_{net}= m_{B}g(1-[tex]\mu[/tex])so the acceleration of block B would be g(1-[tex]\mu[/tex]), and plugging in the values I got the answer -4.34m/s

^{2}rounded to three figures. This would also be the acceleration of block C.Next I needed to find the time it took block C to hit the ground.

y = y

_{0}+ v_{0}_{y}t + 1/2 at^{2}so

y = 1/2 at

^{2}t = sqrt(2y/a)

t = sqrt(2*-1.05m/-4.34m/s

^{2})t = 0.695s rounded to three figures

Did I do this problem correctly? And if not, can you please show me the correct way of doing a problem like this? Thanks.