An 8.0-kg block is attached to a spring with a constant of 2.0 N/m.If the spring is stretched 3.0 m from its equilibrium position and released from rest,the maximum velocity attained by the mass is
A)0.75 m/s
B)1.5 m/s
C)3.0 m/s
D)4.2 m/s
E)15 m/s

Question

Quizplus · Accepted Answer

The maximum velocity can be found using the conservation of energy principle. The potential energy stored in the spring at maximum stretch is converted into kinetic energy at the point of maximum velocity. The potential energy in the spring (PE) is given by $PE = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. The kinetic energy (KE) is given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity.Given: $k = 2.0 \, \text{N/m}$, $x = 3.0 \, \text{m}$, and $m = 8.0 \, \text{kg}$.Calculating the potential energy at maximum stretch:$$PE = \frac{1}{2} \cdot 2.0 \cdot (3.0)^2 = 9.0 \, \text{J}$$Setting this equal to the kinetic energy to find the maximum velocity:$$9.0 = \frac{1}{2} \cdot 8.0 \cdot v^2$$Solving for $v$:$$v^2 = \frac{9.0 \times 2}{8.0} = \frac{18}{8} = 2.25$$$$v = \sqrt{2.25} = 1.5 \, \text{m/s}$$Therefore, the maximum velocity attained by the mass is 1.5 m/s.

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