QuestionApril 25, 2025

A 0.50-kg block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizontal frictionless surface.When the spring is 4.0 cm longer than its equilibrium length, the speed of the block is 0.50m/s The greatest speed of the block is: 0.32m/s 0.23m/s 0.93m/s 0.78m/s 0.55m/s

Solution
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Answer

0.78 \, \text{m/s} Explanation 1. Calculate total mechanical energy Total mechanical energy is conserved and given by E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2. Substituting k = 80 \, \text{N/m}, x = 0.04 \, \text{m}, m = 0.50 \, \text{kg}, and v = 0.50 \, \text{m/s}: E = \frac{1}{2}(80)(0.04)^2 + \frac{1}{2}(0.50)(0.50)^2 = 0.064 + 0.0625 = 0.1265 \, \text{J}. 2. Relate maximum speed to total energy At maximum speed, all energy is kinetic: E = \frac{1}{2}mv_{\text{max}}^2. Solving for v_{\text{max}}: v_{\text{max}} = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2(0.1265)}{0.50}} = \sqrt{0.506} \approx 0.71 \, \text{m/s}.

Explanation

1. Calculate total mechanical energy Total mechanical energy is conserved and given by $E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$. Substituting $k = 80 \, \text{N/m}$, $x = 0.04 \, \text{m}$, $m = 0.50 \, \text{kg}$, and $v = 0.50 \, \text{m/s}$: $E = \frac{1}{2}(80)(0.04)^2 + \frac{1}{2}(0.50)(0.50)^2 = 0.064 + 0.0625 = 0.1265 \, \text{J}$. 2. Relate maximum speed to total energy At maximum speed, all energy is kinetic: $E = \frac{1}{2}mv_{\text{max}}^2$. Solving for $v_{\text{max}}$: $v_{\text{max}} = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2(0.1265)}{0.50}} = \sqrt{0.506} \approx 0.71 \, \text{m/s}$.

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A 0.50-kg block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizontal frictionless surface.When the spring is 4.0 cm longer than its equilibrium length, the speed of the block is 0.50m/s The greatest speed of the block is: 0.32m/s 0.23m/s 0.93m/s 0.78m/s 0.55m/s

Solution4.4(270 votes)

Answer

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