Two blocks are arranged at the ends of a massless string as shown in the figure. The system starts from rest. When the 4.36 kg mass has fallen through 0.332 m, its down- ward speed is 1.25 m/s. The acceleration of gravity is 9.8 m/s².

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Two blocks are arranged at the ends of a
massless string as shown in the figure. The
system starts from rest. When the 4.36 kg
mass has fallen through 0.332 m, its down-
ward speed is 1.25 m/s.
The acceleration of gravity is 9.8 m/s².
5.77 kg
N.
fl
4.36 kg
What is the frictional force between the
5.77 kg mass and the table?
Answer in units of N. Answer in units of
Your response...
i
Previous Responses
X #1.0.526
PALETTE
Transcribed Image Text:Two blocks are arranged at the ends of a massless string as shown in the figure. The system starts from rest. When the 4.36 kg mass has fallen through 0.332 m, its down- ward speed is 1.25 m/s. The acceleration of gravity is 9.8 m/s². 5.77 kg N. fl 4.36 kg What is the frictional force between the 5.77 kg mass and the table? Answer in units of N. Answer in units of Your response... i Previous Responses X #1.0.526 PALETTE
Expert Solution
Step 1

Given m=4.36kg, M=5.77kg.

Let a be the acceleration of the system.

From the FBD of mass m,

T - mg=- ma(negative sign of acceleration is due to the block moved downward).

 =>T=mg-ma..........,(1)

For the block of mass M,

T - f= Ma.......(2)

substitute the value of T from (1) in (2)

mg - ma -f = Ma

solving for force of friction 

f=mg - (m+M)a.......(3)

 

 

 

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