A 0.39-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant)is $570 \mathrm {~N} / \mathrm { m } .$ The block is pulled from its equilibrium position at x = 0.000 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the position of the block is $x = 0.057 \mathrm {~m} ,$ its kinetic energy is closest to
A)0.90 J.
B)0.84 J.
C)0.95 J.
D)1.0 J.
E)1.1 J.

Question

Quizplus · Accepted Answer

The correct answer is A because the kinetic energy of the block is maximum when it passes through the equilibrium position. The kinetic energy of the block is given by:$$K = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2A^2\sin^2(\omega t)$$where m is the mass of the block, v is the velocity of the block, \omega is the angular frequency of the motion, A is the amplitude of the motion, and t is the time.At the position $x = 0.057 \mathrm {~m} ,$ the block is moving with a velocity of:$$v = \omega A\cos(\omega t) = \sqrt{\frac{k}{m}}A\cos(\omega t)$$where k is the spring constant.Substituting this expression for v into the equation for K, we get:$$K = \frac{1}{2}m\left(\sqrt{\frac{k}{m}}A\cos(\omega t)\right)^2 = \frac{1}{2}kA^2\cos^2(\omega t)$$At the position $x = 0.057 \mathrm {~m} ,$ the block is at a displacement of:$$x = A\cos(\omega t) = 0.057 \mathrm {~m}$$Solving for $\cos(\omega t)$, we get:$$\cos(\omega t) = \frac{x}{A} = \frac{0.057 \mathrm {~m}}{0.080 \mathrm {~m}} = 0.7125$$Substituting this value into the equation for K, we get:$$K = \frac{1}{2}kA^2\cos^2(\omega t) = \frac{1}{2}(570 \mathrm {~N} / \mathrm { m})(0.080 \mathrm {~m})^2(0.7125)^2 = 0.90 \mathrm {~J}$$Therefore, the kinetic energy of the block is closest to 0.90 J when the position of the block is $x = 0.057 \mathrm {~m} .$

A 039-Kg Block on a Horizontal Frictionless Surface Is Attached to to an Ideal