NCERT Solution
Class 11
Physics
Laws of Motion
a monkey of mass 40 kg climbs on a rope fig 5 20...

Question:
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/s²upward

UsingNewton's 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 > T_max, the rope will break in this case.

Case (b)

Acceleration of the monkey, a= 4 m/s²downward

UsingNewton's 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 < 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.

UsingNewton's second law of motion, we can write the equation of motion as:

T- mg = ma

T- mg = 0

∴T = mg

= 40 x 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

UsingNewton's second law of motion, we can write the equation of motion as:

mg - T= mg

∴T= m(g- g) = 0

Since T < T_max, the rope will not break in this case.

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