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Question: A constant force F = m<sub>2</sub>g/2 is applied on the block of mass m<sub>1</sub> as shown in fig....

A constant force F = m2g/2 is applied on the block of mass m1 as shown in fig. The string and the pulley are light and the surface of the table is smooth. The acceleration of m1 is –

A

m2g2(m1+m2)\frac { m _ { 2 } g } { 2 \left( m _ { 1 } + m _ { 2 } \right) } towards right

B

m2g2(m1−m2)\frac { m _ { 2 } g } { 2 \left( m _ { 1 } - m _ { 2 } \right) } towards left

C

m2g2(m2−m1)\frac { m _ { 2 } g } { 2 \left( m _ { 2 } - m _ { 1 } \right) } towards right

D

m2g2(m2−m1)\frac { m _ { 2 } g } { 2 \left( m _ { 2 } - m _ { 1 } \right) } towards left

Answer

m2g2(m1+m2)\frac { m _ { 2 } g } { 2 \left( m _ { 1 } + m _ { 2 } \right) } towards right

Explanation

Solution

m1 : T – F = m1a

F = m2 g/2

\ T = m2 g2\frac { m _ { 2 } \mathrm {~g} } { 2 } + m1a .....(i)

m2 : m2g – T = m2a ̃ T = m2g – m2a .... (ii)

Q (i) = (ii)

\ m2 g2\frac { m _ { 2 } \mathrm {~g} } { 2 } + m1a = m2g – m2a ̃ a = m2g2(m1+m2)\frac { m _ { 2 } g } { 2 \left( m _ { 1 } + m _ { 2 } \right) }

short trick

a =  Total force  Total mass \frac { \text { Total force } } { \text { Total mass } } = m2g−Fm1+m2\frac { m _ { 2 } g - F } { m _ { 1 } + m _ { 2 } } = m2g2(m1+m2)\frac { m _ { 2 } g } { 2 \left( m _ { 1 } + m _ { 2 } \right) }