Two light springs (1) and (2) of spring constant k and 2k are connected in series as shown in figure. A block of mass M is hanging vertically in equilibrium. If Δx is the elongation in spring (2) in cm, then value of Δx will be (Given, Mgk = 8 cm)Given, 𝑀𝑔𝑘 = 8 cm
Question
Two light springs (1) and (2) of spring constant k and 2k are connected in series as shown in figure. A block of mass M is hanging vertically in equilibrium. If Δx is the elongation in spring (2) in cm, then value of Δx will be (Given, Mgk = 8 cm)Given, 𝑀𝑔𝑘 = 8 cm
Solution
The system is in equilibrium, so the total force acting on the block is zero. This means that the weight of the block (Mg) is balanced by the total force exerted by the springs.
The total force exerted by the springs is the sum of the forces exerted by each spring. The force exerted by a spring is given by Hooke's law, F = kx, where k is the spring constant and x is the displacement from the equilibrium position.
For spring 1, the force is F1 = kx1. For spring 2, the force is F2 = 2kx2.
The total force is F = F1 + F2 = kx1 + 2kx2.
Since the system is in equilibrium, the total force is equal to the weight of the block, Mg. Therefore, we have:
Mg = kx1 + 2kx2.
We are given that Mgk = 8 cm. This means that Mg = 8k. Substituting this into the equation gives:
8k = kx1 + 2kx2.
Dividing through by k gives:
8 = x1 + 2x2.
We are asked to find the value of Δx, which is the elongation in spring 2. This is equal to x2. Therefore, we need to solve the equation for x2.
From the equation, we can express x1 in terms of x2:
x1 = 8 - 2x2.
Substituting this into the equation gives:
8 = (8 - 2x2) + 2x2.
Solving for x2 gives:
x2 = 8/3 cm.
Therefore, the elongation in spring 2, Δx, is 8/3 cm.
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