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The potential energy of an object attached to a spring is 2.80 j at a location where the kinetic energy is 1.50 j. The amplitude a of the simple harmonic motion is 18.0 cm. Calculate the spring constant k and the magnitude of the largest force f_spring,max that the object experiences.

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Final answer:

To find the spring constant, use the formula U = (1/2)kx^2 with the given potential energy and displacement. The spring constant is 43.62 N/m. The maximum force experienced by the object is found using Hooke's law, F = -kx, with the given displacement.

Step-by-step explanation:

To find the spring constant, we can use the formula for potential energy of a spring, U = (1/2)kx^2, where U is the potential energy, k is the spring constant, and x is the displacement from equilibrium. In this case, we are given that the potential energy is 2.80 J and the displacement is the amplitude, a, which is 18.0 cm. Plugging these values into the formula, we have 2.80 J = (1/2)k(0.18 m)^2. Simplifying the equation, we find k = 43.62 N/m.

To find the maximum force experienced by the object, we can use Hooke's law, which states that F = -kx, where F is the force, k is the spring constant, and x is the displacement from equilibrium. Since the object is at its maximum displacement at the amplitude, the maximum force is F = -k(0.18 m) = -43.62 N/m * 0.18 m = -7.85 N or 7.85 N, depending on the direction of the force.

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