Question

a constant force F=m2g/2 is applied on the block of mass m1 as shown in figure. if the string and the pulley are light and the surface of the table is smooth,then the acceleration of m1 is
a)m2g/2(m1+m2) towards left
b) m2g/2(m1+m2) towards right
c) m2g/2(m1-m2) towards left
d)m2g/2(m1-m2) towards right​

Answer 1

From the free body diagram (FBD) of block of mass m1

m1a = T - F _[i]

From FBD of block of mass m2

m2a = m2g - T _[ii]

Adding Both the Equations, we get

a (m1 + m2 ) = m2g - m2g / 2 ...[becoz F=m2g]

=> a = m2g / 2(m1 + m2 )

Therefore acceleration of mass m1 ,

a = m2g / 2(m1 + m2 ) Towards the right .

Answer :

b) a=m2g/2(m1 +m2) towards right !

[ See the attachment for better understanding ]

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Answer 2

Answer:

b) is the correct answer..

Hope it will be helpful ☺️ thank you!!