a monkey of mass 20kg climbs on a rope shown in figure which can stand a maximum tension in rope (g=10m/s^2)
(a) climbs up with an acceleration of 6m/s^2 (b) climbs down with an acceleration of 4m/s^2
(c) climbs a with a uniform speed of 5m/s^2
(d) falls down the Rope nearly freely under Gravity?
Answers
Answer:
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/s
2
upward
Using Newtons 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.
(b)
Acceleration of the monkey, a = 4 m/s
2
downward
Using Newtons 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.
(c)
The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a = 0.
Using Newtons 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 < T
max
, the rope will not break in this 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 Newtons 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.