The pulley arrangements of the figure. (a) and (b) are identical. The mass of the rope is negligible. In (a) the mass m is lifted up by attaching a mass 2m to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 m g . The acceleration of m in both cases is?

Question

The pulley arrangements of the figure. (a) and (b) are identical. The mass of the rope is negligible. In (a) the mass m is lifted up by attaching a mass 2m to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 m g . The acceleration of m in both cases is? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-8xp1lzhkpwtix4ad.jpg" /> (A) g/3 , g/3 (B) g/3 , g (C) g/4 , g/3 (D) g/4 , g

Tardigrade Admin · Accepted Answer

Case A For mass m T - m g = m a (i) For mass 2m 2 m g - T = 2 m a (ii) Solution

From (i) and (ii), we get – a = (g/3) …(iii) Case B T-mg=ma' ⇒ 2 m g - m g = m a ' [. ∵ T = 2 m g ]. ∴ a ' = g …(iv) Solution

$g /3$ , $g /3$

$g /3$ , $g$

$g /4$ , $g /3$

$g /4$ , $g$

Case A
For mass $m$
$T - m g = ma$ (i)
For mass $2 m$
$2 m g - T = 2 ma$ (ii)

From (i) and (ii), we get –
$a = \frac{g}{3}$ …(iii)
Case B
$T - m g = m a^{'}$
$\Rightarrow 2 m g - m g = m a^{'}$ $[∵ T = 2 m g]$
$∴ a^{'} = g$ …(iv)

g/3 , g/3

g/3 , g

g/4 , g/3

g/4 , g

Case A For mass m T−mg=ma (i) For mass 2m 2mg−T=2ma (ii) From (i) and (ii), we get – a=3g​ …(iii) Case B T−mg=ma′ ⇒2mg−mg=ma′ [∵T=2mg] ∴a′=g …(iv)

$g /3$ , $g /3$

$g /3$ , $g$

$g /4$ , $g /3$

$g /4$ , $g$

Case A
For mass $m$
$T - m g = ma$ (i)
For mass $2 m$
$2 m g - T = 2 ma$ (ii)

From (i) and (ii), we get –
$a = \frac{g}{3}$ …(iii)
Case B
$T - m g = m a^{'}$
$\Rightarrow 2 m g - m g = m a^{'}$ $[∵ T = 2 m g]$
$∴ a^{'} = g$ …(iv)