Question

Two men of masses $m_{1}and{m}_2$ hold on the opposite ends of a rope passing over a frictionless pulley. The mass $m_1$ climbs up the rope with an acceleration of $1.2m/s^2$ relative to the rope. The man $m_2$ climbs up the rope with an acceleration of $2.0m/s^2$ relative to the rope. They start from rest and are initially separated by $5m$. (Given $m_1=40kg$ and $m_2=60kg$)

• A. Tension in the rope is $555.68N$
• B. Tension in the rope is $556.8N$
• C. Time after which they will be at same horizontal level is $1.47sec$
• D. Time after which they will be at same horizontal level is $2.94sec$

Answer 1

Answer: (B), (C)

Tension in the rope is $556.8N$
Time after which they will be at same horizontal level is $1.47sec$

assuming pulley is rotating clockwise with an acceleration $a$.

also $a_{mg}=a_{mr}+a_{rg}$, so acceleration of $m_1$ is $a+1.2$ and acceleration of $m_2$ is $(-a+2)$ assuming both in upward direction.

writing equation of motion

$T-600=60(-a+2)$

and $T-400=(a+1.2)$

solving above equations we get $a=2.72$ in $clockwise$ direction.

also tension in cable $T=556.8N$

now net acceleration of both the man relative to ground is $m_1=3.84m/s^{2}upwards$ and $m_2=0.72m/s^{2}downwards$

acceleration of $m_2$ wrt $m_1$ is $4.56m/s^{2}upwards$ and distance between both is $5m$.

using equation $S=ut+\frac{1}{2}a{t}^2$

where $u=0,a=4.56$ we get $T=1.48sec$.

so best possible options are B,C.