Q. 264.3( 6 Votes )

A constant force F=m2g/2 is applied on the block of mass m1 as shown in figure (5-E 10). The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.

Answer :

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) = m2g - F


The acceleration of mass m1, a= , towards the right.

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