Question

7. A monkey of mass 40 kg climbs on a rope 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 6ms 2
(b) Climbs down with an acceleration of 4ms?
(C) Climbs up with a uniform speed of 5ms -1
(d) Falls down the rope nearly freely under gravity?

Answer 1

Answer:

A) Climbs up with an acceleration of 6m/s2

Explanation:

(a)

Mass of the monkey, m = 40 kg

Acceleration due to gravity, g = 10 m/s

Maximum tension that the rope can bear, Tmax = 600 N

Acceleration of the monkey, a = 6 m/s2 upward

Using Newtons second law of motion, we can write the equation of motion as:

T  mg = ma

T = m(g + a)

= 40 (10 + 6) 

= 640 N

Since T > Tmax,

the rope will break in this case.

(b)

Acceleration of the monkey, a = 4 m/s2 downward

Using Newtons second law of motion, we can write the equation of motion as:

mg  T = ma

T = m (g- a)

= 40(10-4) 

= 240 N

Since T < Tmax,

the rope will not break in this case.

(c)

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

Using Newtons second law of motion, we can write the equation of motion as:

T - mg = ma

T- mg = 0 

T = mg

= 40 × 10 

= 400 N

Since T < Tmax,

the rope will not break in this 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 Newtons second law of motion, we can write the equation of motion as:

mg + T = mg

T = m(g -g) = 0

Since T < Tmax,

the rope will not break in this case.