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It takes a force of 36,340lb to compress a coil spring assembly from its free height of 11 inches to its fully compressed height of 6 inches.

a. What is the assembly's force constant?
b. How much work does it take to compress the assembly the first half inch?

User Digvijay
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Final answer:

The force constant of the spring assembly is approximately 286,299.2 N/m. The work required to compress the assembly by the first half inch is approximately 0.2291 J.

Step-by-step explanation:

To find the force constant of the spring assembly, we can use Hooke's Law which states that the force exerted by a spring is directly proportional to its displacement from equilibrium. Using the formula F = kx, where F is the force applied, k is the force constant, and x is the displacement, we can rearrange the equation to solve for k. In this case, the force is 36,340 lb and the displacement is 11 - 6 = 5 inches. Converting the force to Newtons (1 lb = 4.4482 N) and the displacement to meters (1 inch = 0.0254 m), we can calculate the force constant:

F = kx

36,340 lb = k(5 inches)

36,340 lb = k(0.127 m)

k = 36,340 lb / 0.127 m

k ≈ 286,299.2 N/m

b. To find the work required to compress the assembly by the first half inch, we can use the formula for work done on a spring, which is given by W = (1/2)kx^2. In this case, the force constant is 286,299.2 N/m and the displacement is 0.5 inches. Converting the displacement to meters, we can calculate the work:

W = (1/2)kx^2

W = (1/2)(286,299.2 N/m)(0.0127 m)^2

W ≈ 0.2291 J

User Salim Ibrohimi
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