Question

A monkey climbs up and another monkey climbs down a rope hanging from a tree with same uniform acceleration separately. If the respective masses of monkeys are in the ratio 2 : 3, the common acceleration must be

• A. $\frac{g}{5}$
• B. $6g$
• C. $\frac{g}{2}$
• D. $g$

Answer 1

Answer: (A)

Let $T$ be the tension of a rope. When monkey climbs up with acceleration $a,$
$T - m_1g = m_1a ...(i)$
When another monkey climbs down with same acceleration a, then
$m_2g - T = m_2a ...(ii)$
Adding (i) and (ii), we get
or $\left(1-\frac{m_{1}}{m_{2}}\right)g=\left(\frac{m_{1}}{m_{2}}+1\right)a$ or $\left(1-\frac{2}{3}\right)g=\left(\frac{2}{3}+1\right)a$
$\frac{1}{3}g=\frac{5}{3}a$
or $a=\frac{g}{5}$