Question

A constant force F = m2g/2 is applied on the block of mass m1 as shown in figure (5−E10). The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.

Answer 1

Given :

From the free-body diagram of block of mass m1,

m1a = T − F    …(i)

From the free-body diagram of block of mass m2,

m2a = m2g − T    …(ii)

Adding both the equations, we get:

a(m1+m2)=m1g-m2g/2          

∴F=m2g/2

a=m2g/ 2(m1+m2)

∴The acceleration of mass m1,  a=m2g/2m1+m2, towards the right.

Answer 2

Answer:

Refer to the attachment for your answer