A 3.7-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant)is $450 \mathrm {~N} / \mathrm { m } .$ The block is pulled from its equilibrium position at x = 0.000 m to a position x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. The maximum elastic potential energy of the system is closest to
A)1.4 J.
B)1.3 J.
C)1.6 J.
D)1.7 J.
E)1.8 J.

Question

Quizplus · Accepted Answer

The maximum elastic potential energy of the system is equal to the work done in stretching the spring from its equilibrium position to its maximum displacement. The work done is given by:$$W = \frac{1}{2}kx^2$$where k is the spring constant and x is the displacement from equilibrium. Substituting the given values, we get:$$W = \frac{1}{2}(450 \mathrm {~N} / \mathrm { m })(0.080 \mathrm {~m})^2 = 1.44 \mathrm {~J}$$Therefore, the maximum elastic potential energy of the system is closest to 1.4 J.

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