In an Atwood’s machine two bodies having masses 10kg and 15kg are suspended. After it has been in motion for 5 seconds, 10 kg mass is removed from heavier body. Further how much time (in second) will elapse before the directions of motion of the bodies are reversed. (g = 10 m/s2)
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11th
Physics
Laws of Motion
Application of Newton's Laws of Motion
Two bodies of masses 10 kg...
PHYSICS
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Asked on November 22, 2019 by
Akinapally Chatterjee
Two bodies of masses 10 kg and 20 kg respectively, kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to(i) A (ii) B along the direction of string. What is the tension in the string in each case?
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ANSWER
Horizontal force,F=600 N
Mass of body A, m
1
=10 kg
Mass of body B, m2=20 kg
Total mass of the system, m=m1+m2=30 kg
Using Newtons second law of motion, the acceleration (a) produced in the system can be calculated as:
F=ma
a=
m
F
=
30
600
=20m/s
2
When force F is applied on body A:
The equation of motion can be written as: F−T=m
1
a
T=F−(m
1
)a
=600−10×20=400 N (i)
When force F is applied on body B:
The equation of motion can be written as:
F−T=m
2
a
T=F−(m
2
)a
T=600−20×20=200 N (ii)
which is different from value of T in case (i). Hence our answer depends on which mass end, the force is applied.
solution
Answer:
given = 2 bodies having mass
m1 = 10Kg
m2 = 15Kg
tension (T2) = 5sec
g = 10
to find = time (in second) will elapse before the directions of motion of the bodies are reversed ? (T1) tension
solution =
we know that,
T2 = m2g
T2= 0.3kg x 10m/s²............... ( 15/5 = 3 )
T2= 3N
T1 =m1g +T2
T1 = 0.2 Kg x 10m/s²+ 3............... ( 10/5=2)
T1 = 5N
above two equations are according to block diagrams.
so we can conclude that the tension in the 2 string are 5N and 3N.
