Two masses of 8kg and 12kg connected at the two ends of a light inextensible string passes over a frictionless pulley find acceleration of the masses and tension in the string . when the masses are released
Answers
The tension in the string is 96 N.
Explanation:
Given data:
- Smaller mass, m1 = 8 kg
- Larger mass, m2 = 12 kg
- Tension in the string = T
Applying Newton’s second law of motion to the system of each mass:
For mass m1:
The equation of motion can be written as:
T – m1g = ma … (1)
For mass m2:
The equation of motion can be written as:
m2g – T = m2a … (2)
Adding equations (i) and (ii), we get:
(m2 - m1)g = (m1 + m2)a
a = ( (m2 - m1) / (m1 + m2) )g ....(3)
= (12 - 8) / (12 + 8) × 10 = 4 × 10 / 20 = 2 ms-2
Therefore, the acceleration of the masses is 2 m/s^2.
Substituting the value of a in equation (ii), we get:
m2g - T = m2(m2 - m1)g / (m1 + m2)
T = (m2 - (m2^2 - m1m2) / (m1 + m2) )g
= 2m1m2g / (m1 + m2)
= 2 × 12 × 8 × 10 / (12 + 8)
= 96 N
Therefore, the tension in the string is 96 N.
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Two masters of 10 kg and 20 kg respectively are connected by a massless spring. A force of 200 Newton acts on the 20 kg mass. At the instant shown, the 10 kg mass has an acceleration of 12 metre per second square then what is the acceleration of 20 kg mass ?
https://brainly.in/question/1616200
Answer:
- Tension (T) in the string is 96 N.
Given:
- Mass of lighter block (M₁) = 8 Kg
- Mass of Higher block (M₂) = 12 kg.
- Let the Tension be " T "
- Let the Acceleration be " a "
Explanation:
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Since, M₂ is Heavier ( 12 Kg > 8 Kg ) so it will pull the system Downwards.
# Refer the Attachment for Figure.
Now, Applying Law of Equilibrium For 8 Kg Block.
Equation will be,
⟼ T - M₁g = Ma
Substituting the Values,
⟼ T - 8 x 10 = 8 x a
⟼ T - 80 = 8 a
⟼ T = 8 a + 80 ___ [1]
Now, Applying Law of Equilibrium For 12 Kg Block.
Equation will be,
⟼ M₂g - T = M₂a
Substituting the Values,
⟼ 12 x 10 - T = 12 x a
⟼ 120 - T = 12 a
⟼ 120 - T = 12 a ____ [2]
Adding [1] & [2]
⟼ T + 120 - T = 8 a + 80 + 12 a
⟼ 120 = 20 a + 80
⟼ 120 - 80 = 20 a
⟼ 20 a = 120 - 80
⟼ 20 a = 40
⟼ a = 40 / 20
⟼ a = 2 m/s²
⟼ a = 2 m/s².
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Substituting the value in Equation [1]
⟼ T = 8 a + 80 ___ [1]
Now,
⟼ T = 8 x 2 + 80
⟼ T = 16 + 80
⟼ T = 96 N
⟼ T = 96 N
∴ Tension (T) in the string is 96 N.
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