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Standard XII

Physics

Net Contact Force

Two masses ...

Question

Two masses $m_{1}$ and $m_{2}$ are tied with a thread passing over a pulley, $m_{1}$ is on a frictionless horizontal surface and $m_{2}$ is hanging freely. If acceleration due to gravity is $g$ , the acceleration of $m_{1}$ in this arrangement will be:

g

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g / (m_{1} + m_{2})

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g / m

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g / (m_{1} - m_{2})

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Solution

The correct option is D $g / (m_{1} + m_{2})$
Equation of motion-
$T = m_{1} a$
$m_{2} g - T = m_{2} a$
$\Rightarrow$ $a = \frac{m_{2} g}{m_{1} + m_{2}}$

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Two masses m1 and m2 are tied with a thread passing over a pulley, m1 is on a frictionless horizontal surface and m2 is hanging freely. If acceleration due to gravity is g, the acceleration of m1 in this arrangement will be:

The correct option is D g/(m1+m2)Equation of motion-T=m1am2g−T=m2a⇒ a=m2gm1+m2

Two masses $m_{1}$ and $m_{2}$ are tied with a thread passing over a pulley, $m_{1}$ is on a frictionless horizontal surface and $m_{2}$ is hanging freely. If acceleration due to gravity is $g$ , the acceleration of $m_{1}$ in this arrangement will be:

The correct option is D $g / (m_{1} + m_{2})$
Equation of motion-
$T = m_{1} a$
$m_{2} g - T = m_{2} a$
$\Rightarrow$ $a = \frac{m_{2} g}{m_{1} + m_{2}}$