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A Monkey of Mass 40 Kg Climbs on a Rope in Given Figure Which Can Stand a Maximum Tension of 600 N. in Which of the Following Cases Will the Rope Break: the Monkey Climbs up with an Acceleration of 6 M Sclimbs Down with an Acceleration of 4 M - Physics

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Question

A monkey of mass 40 kg climbs on a rope in given Figure 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).

Solution 1

Case (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 Newton’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 > Tmax, the rope will break in this case.

Case (b)

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

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

mg – ma

T = (g – a)

= 40(10 – 4)

= 240 N

Since T < Tmax, 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= ma

T – mg = 0

mg

= 40 × 10

= 400 N

Since T < Tmax, 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:

mg – T = mg

T = m(g – g) = 0

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

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Solution 2

(a) When the monkey climbs up with an acceleration a, then T – mg = ma where T represents the tension in figure.

:. T = mg + ma = m (g +a)

or T = 40 kg(10 + 6) ms^(-2) = 640 N

But the rope can withstand a maximum tension of 600 N. So the rope will break

b) When the monkey is climbing down with an acceleration, then

mg - T =ma (Figure b)

=> T = mg - ma = m(g -a)

or T = 40 kg x (10 - 4) `ms^2` = 240 N

(c) When the monkey climbs up with uniform speed, then T mg = 40 kg x 10 ms-2 = 400 N The rope will hot break.

(d) When the monkey is falling freely, it would be a state of weightlessness. So, tension will be zero and the rope will not break.

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Chapter 5: Laws of Motion - Exercises [Page 113]

APPEARS IN

NCERT Physics [English] Class 11
Chapter 5 Laws of Motion
Exercises | Q 33 | Page 113

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