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a spring has a equilibrium length of 0.100 m. when a force of 40.0 n is applied to the spring, the spring has a length of 0.140 m. what is the value of the spring constant of this spring?

User Sakiboy
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The value of the spring constant of this spring will be 50,000 N/m, which has a equilibrium length of 0.100m applied by 40N force.

Step-by-step explanation:

Formula for the spring force is given as- F = (1/2) kx²

where, F = force, k= spring constant and x = change in length of the spring.

Change in length of the spring = Changed length - equilibrium length

Change in length of the spring(x) = 0.140m - 0.100m = 0.040m Putting the values as F = 40.0 N, x = 0.040m, k =?

F = (1/2) kx²40 = (1/2) × k × (0.040)²k = 80/0.0016k = 50,000 N/m


Therefore, the value of the spring constant of this spring will be 50,000 N/m

User Jomel Imperio
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