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Asked by akp70528438 | 28 Nov, 2021, 10:05: AM
Figure shows the forces acting on blocks of masses m1 and m2 . Weight m1g is acting downward. Equal reaction force R is acting upward
when the block of mass m1 is moving , friction force μR is acting against the motion.Tension force T is acting along the string.

If we apply Newton's law to the block of mass m1 , we get

T - μ R  = T - ( μ m1 g ) = m1 a  ...............................(1)

where a is acceleration of system.

f we apply Newton's law to the block of mass m2 , we get

m2 g - T = m2 a ...............................(2)

By adding eqn.(1) and (2) , we get ,

m2 g - ( μ m1 g )  = ( m1 + m2 ) a

Hence acceleration a  is given from above expression as

......................................(3)

We get Tension force T fromm eqn.(2)  as

T = m2 ( g - a )   .............................(4)

By substituting acceleration a from eqn.(3) , we get tension force T from eqn.94) as

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