Question

A monkey of mass $12\text{kg}$ climbs up a light rope as shown in figure. The rope passes over a light and frictionless pulley and it is attached to a bunch of bananas having mass $16\text{kg}$. What will be maximum acceleration of monkey without lifting the bunch of bananas? (Take $g=10{\text{m/s}}^2$)

• A. $2.33{\text{m/s}}^2$
• B. $3.33{\text{m/s}}^2$
• C. $4.33{\text{m/s}}^2$
• D. $5.33{\text{m/s}}^2$

Answer 1

The correct option is B $3.33{\text{m/s}}^2$

Mass of monkey ($m$)= $12\text{kg}$
Mass of banana bunch ($M$)= $16\text{kg}$
From FBD of banana bunch:


$\Rightarrow$The banana bunch will just leave the ground when $N=0$
Applying equilibrium condition in vertical direction and putting $N=0$ for just leaving the contact surface

$T+N=Mg$
$\Rightarrow T=Mg$
$\therefore T=16\times 10=160\text{N}.........\left(i\right)$

From F.B.D. of monkey, the maximum acceleration will correspond to $T=160\text{N}$ in string.


From Newton's $2^{\text{nd}}$ law:
$T-mg=ma$
$T-12g=12\times a$
$\Rightarrow T-120=12\times a...........\left(ii\right)$

By equation $\left(i\right)$ and $\left(ii\right)$:
$160-120=12\times a$
$\therefore a=\frac{40}{12}=3.33{\text{m/s}}^2$
This is the maximum acceleration of monkey for which banana bunch will not be lifted.