### Q.33:- A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

(a) climbs up with an acceleration of 6 m s^{–2}

(b) climbs down with an acceleration of 4 m s^{–2}

(c) climbs up with a uniform speed of 5 m s^{–1}

(d) falls down the rope nearly freely under gravity?

*(Ignore the mass of the rope).*

**Answer:-**

__Case (a)__

Mass of the monkey, *m* = 40 kg

Acceleration due to gravity, *g* = 10 m/s

Maximum tension that the rope can bear, *T*_{max} = 600 N

Acceleration of the monkey, *a* = 6 m/s^{2} upward

Using Newton’s second law of motion, we can write the equation of motion as:

*T* – *m*g* = ma*

∴*T* = *m*(g* + a*)

= 40 (10 + 6)

= 640 N

Since *T* > *T*_{max}, the rope will break in this case.

__Case (b)__

Acceleration of the monkey, *a* = 4 m/s^{2} downward

Using Newton’s second law of motion, we can write the equation of motion as:

*m*g – *T *= *ma*

∴*T* = *m *(g* – a*)

= 40(10 – 4)

= 240 N

Since *T* < *T*_{max}, the rope will not break in this case.

__Case (c)__

The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., *a* = 0.

Using Newton’s second law of motion, we can write the equation of motion as:

*T* – *m*g *= ma*

*T* – *m*g = 0

∴*T *= *m*g

= 40 × 10

= 400 N

Since *T* < *T*_{max}, the rope will not break in this case.

__Case (d)__

When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e., *a = *g

Using Newton’s second law of motion, we can write the equation of motion as:

*m*g – *T* =* m*g

∴*T* = *m*(g* – *g) = 0

Since *T* < *T*_{max}, the rope will not break in this case.