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We have the two masses, m₁ and m₂. The applied force is F, k is the spring constant, X is the extension of the spring.
Then, the total mass of the system is (assuming spring to be massless):
M = m₁ + m₂ = 2 kg + 4 kg = 6 kg
Now using Newton's second law, we can find the acceleration of the system as:
a = F / (m₁ + m₂) = 10 N / 6 kg = 5/3
Hence the tension in the spring would be given by:
T = m₁×a = 2 kg × 5/3 = 10/3
Therefore, by Hooke’s law, we can find the extension of the spring.
F = k×X
Since F = T, so
m₁ × F/(m₁ + m₂) = k×X
So, X = m₁×F / [k×(m₁ + m₂)]
= 10/3 N
Then, the total mass of the system is (assuming spring to be massless):
M = m₁ + m₂ = 2 kg + 4 kg = 6 kg
Now using Newton's second law, we can find the acceleration of the system as:
a = F / (m₁ + m₂) = 10 N / 6 kg = 5/3
Hence the tension in the spring would be given by:
T = m₁×a = 2 kg × 5/3 = 10/3
Therefore, by Hooke’s law, we can find the extension of the spring.
F = k×X
Since F = T, so
m₁ × F/(m₁ + m₂) = k×X
So, X = m₁×F / [k×(m₁ + m₂)]
= 10/3 N
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