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Two blocks of masses m1 and m2 are conne...


Two blocks of masses `m_1` and `m_2` are connected as shown in the figure. The acceleration of the block `m_2` is :

A

`(m^2g)/(m_1+m_2)`

B

`(m_1g)/(m_1+m_2)`

C

`(4m_2g-m_1g)/(m_1+m_2)`

D

`(m_2g)/(m_1+4m_2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Let `a=acc^n` of `m_1acc^n` of pulley `=(a+0)/(2)=(a)/(2)`
If acceleration of `m_2=b`
Then `0+(b)/(2)=(a)/(2)`
Hence `a=b`
`T=m_1a`,`m_2g-T=m_2a`
`a=(m_2g)/(m_1+m_2)`
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