If the pulley is massless and moves with an upward acceleration a 0- find the acceleration of block m 1 w-r-t elevator-A- m2-m1 a0-m1C- m2-m1g-a0-m1-m2D- m2-m1 g-m1

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Acceleration

If the pulley...

Question

If the pulley is massless and moves with an upward acceleration $a_{o}$ , find the acceleration of block $m_{1}$ w.r.t elevator.

\frac{(m_{2} - m_{1}) (g - a_{o})}{(m_{1} + m_{2})}

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\frac{(m_{2} - m_{1}) (g + a_{o})}{(m_{1} + m_{2})}

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\frac{(m_{2} - m_{1}) g}{m_{1}}

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\frac{(m_{2} - m_{1}) a_{o}}{m_{1}}

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Solution

The correct option is B $\frac{(m_{2} - m_{1}) (g + a_{o})}{(m_{1} + m_{2})}$

Let the acceleration of block $m_{1}$ w.r.t the pulley be $a$ (upward) and the acceleration of $m_{2}$ w.r.t the pulley be $a$ (downward).
$∴$ Equation for $m_{1}$ :
$T - m_{1} g = m_{1} a_{1}$
where $a_{1} = a_{0} + a$
So, $T - m_{1} g = m_{1} (a_{0} + a) . . . . . (1)$

Now, equation for $m_{2}$ :
$T - m_{2} g = m_{2} a_{2}$
where $a_{2} = (a_{0} - a)$
Therefore, $T - m_{2} g = m_{2} (a_{0} - a) . . . . . . (2)$

Solving both equations $(1)$ & $(2)$ , we get
$m_{1} (a_{0} + a) + m_{1} g - m_{2} g = m_{2} (a_{o} - a)$
$\Rightarrow a (m_{1} + m_{2}) = (m_{2} - m_{1}) (g + a_{o})$
$\Rightarrow a = [\frac{m_{2} - m_{1}}{m_{1} + m_{2}}] (g + a_{0})$

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Acceleration

Standard XII Physics

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If the pulley is massless and moves with an upward acceleration ao, find the acceleration of block m1 w.r.t elevator.

If the pulley is massless and moves with an upward acceleration $a_{o}$ , find the acceleration of block $m_{1}$ w.r.t elevator.