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A constant force F=m2g/s is applied on t...

A constant force `F=m_2g/s` is applied on the block of mass `m_1` as shown in figure. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of `m_1`

Text Solution

Verified by Experts

Given Reset as
Here `sintheta_1 = 4/5`
We know `g sin theta_1 = a+T` …………i ltbgt Again , `sin theta_2 = 3/5`
We know, `T-gsin theta_2 -a=0`
`rarr T=g sin theta_2 +a` ……..ii.

From equations i. and ii.
`g sin theta_2+a+a-g sin theta_1 =0`
`rarr 2a=gsin theta_1 - g sin theta_2 `
` rarr a= g/5 xx 1/2`
`= g/10`
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Knowledge Check

  • Two blocks of masses m and 2m are placed on a smooth horizontal surface as shown. In the first case only a force F_(1) is applied from the left. Later only a force F_(2) is applied from the right. If the force acting at the interface of the two blocks in the two cases is same, then F_(1):F_(2) is

    A
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    B
    `1:2`
    C
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    D
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