A 2.2-kg object is suspended from a spring with k = 18 N/m. The mass is pulled 0.35 m downward from its equilibrium position and allowed to oscillate. What is the maximum kinetic energy of the object?
A)0.25 J
B)0.50 J
C)1.1 J
D)2.0 J
E)4.0 J

Question

Quizplus · Accepted Answer

We can use conservation of energy to find the maximum kinetic energy of the object. At the maximum displacement, all of the potential energy stored in the spring is converted to kinetic energy. Therefore, we can equate the initial potential energy to the final kinetic energy:
(1/2)kx_max^2 = (1/2)mv_max^2
where x_max = 0.35 m and v_max is the maximum velocity of the object.
Solving for v_max, we get:
v_max = sqrt(kx_max^2/m) = sqrt((18 N/m)(0.35 m)^2/2.2 kg) ≈ 1.1 m/s
Therefore, the maximum kinetic energy is:
(1/2)mv_max^2 = (1/2)(2.2 kg)(1.1 m/s)^2 ≈ 1.1 J
The closest choice is C, which is the correct answer.

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