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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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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