Newton’s Second Law corrected the old belief that force is needed to maintain motion. This idea came from the philosopher ______.
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प्रश्न
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).
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उत्तर १
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.
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उत्तर २
(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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