A 2.0-kg block sliding on a frictionless horizontal surface is attached to one end of a horizontal spring (k = 600 N/m) which has its other end fixed. The speed of the block when the spring is extended 20 cm is equal to 3.0 m/s. What is the maximum speed of this block as it oscillates?

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Quizplus · Accepted Answer

The maximum speed of the block occurs when the spring is at its equilibrium position, where all the energy is kinetic. Using the conservation of energy principle, the total mechanical energy at any point during the oscillation is constant. When the spring is extended by 20 cm (0.2 m) and the block's speed is 3.0 m/s, the mechanical energy is a combination of kinetic energy (due to the block's speed) and potential energy (due to the spring's extension). The kinetic energy (KE) is given by $KE = \frac{1}{2}mv^2$ and the potential energy (PE) stored in the spring is given by $PE = \frac{1}{2}kx^2$, where $m$ is the mass of the block, $v$ is the velocity of the block, $k$ is the spring constant, and $x$ is the extension of the spring from its equilibrium position.Given:- $m = 2.0 \, \text{kg}$- $v = 3.0 \, \text{m/s}$ (when $x = 0.2 \, \text{m}$)- $k = 600 \, \text{N/m}$- $x = 0.2 \, \text{m}$The total mechanical energy when the spring is extended by 20 cm is:$$E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}(2.0)(3.0)^2 + \frac{1}{2}(600)(0.2)^2 = 9 + 12 = 21 \, \text{J}$$At the equilibrium position, all this energy will be kinetic, so:$$E = \frac{1}{2}mv_{\text{max}}^2$$Solving for $v_{\text{max}}$:$$21 = \frac{1}{2}(2.0)v_{\text{max}}^2$$$$v_{\text{max}}^2 = \frac{21}{1}$$$$v_{\text{max}} = \sqrt{21} \approx 4.58 \, \text{m/s}$$The closest answer to $4.58 \, \text{m/s}$ is 4.6 m/s, which is option A.

A 20-Kg Block Sliding on a Frictionless Horizontal Surface Is Attached