2003 AIPMT Laws of MotionNewton's Second Law: Tension in a Rope Easy Question 61 of 72 in Laws of Motion

A monkey of mass 20 kg is holding a vertical rope. The rope will not break when a mass of 25 kg is suspended from it but will break if the mass exceeds 25 kg. What is the maximum acceleration with which the monkey can climb up along the rope ? (g = 10 m/s$^2$)

A 5 m/s$^2$
B 10 m/s$^2$
C 25 m/s$^2$
D 2.5 m/s$^2$

The answer is D.

Solution

Maximum tension the rope can bear: $T_{max} = 25g = 25 \times 10 = 250$ N
When the monkey climbs up with acceleration a, the forces on it are the tension T (upward, the rope's reaction to its pull) and its weight mg (downward).
$T - mg = ma \Rightarrow T = m(g + a)$
For the maximum acceleration: $T_{max} - mg = ma$
$250 - 20 \times 10 = 20a \Rightarrow a = \dfrac{50}{20} = 2.5$ m/s$^2$
(Equivalently $ma = T_{max} - mg = 25g - 20g = 5g$, so $a = \dfrac{5g}{20} = \dfrac{g}{4} = \dfrac{10}{4} = 2.5$ m/s$^2$.)

Concept Maximum tension and acceleration

Laws of Motion in past papers

72 questions from this chapter have appeared across 27 exam years.