  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). Fig. 5.20

Solution

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 – T = ma ∴T = m (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 – mg = ma T – mg = 0 ∴T = 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. PhysicsPart - IStandard XII

Suggest Corrections  0  Similar questions
View More  Same exercise questions
View More  People also searched for
View More 