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). from Physics Laws of Motion Class 11 Manipur Board
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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).


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, equation of motion is

                   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, the equation of motion is, 

                                  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, equation of motion is, 

                     T – m= ma 

                     T – mg = 0
 

∴                           m

                               = 40 × 10  

                               = 400 N 

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

Case (d) 

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

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

                          mg – T = m

∴                                 T = m(g – g) = 0 

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