A 3.3-kg block is attached to a horizontal ideal spring with a spring constant of 181 N/m. When at its equilibrium length, the block attache

Question

A 3.3-kg block is attached to a horizontal ideal spring with a spring constant of 181 N/m. When at its equilibrium length, the block attached to the spring is moving at 5.5 m/s. The maximum amount that the spring can stretch is _____ m. Round your answer to the nearest hundredth.

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RI SƠ 3 years 2021-08-16T03:39:27+00:00 1 Answers 29 views 0

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  1. Gerda
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    2021-08-16T03:40:56+00:00
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    Answer:

    0.74 m

    Explanation:

    Given:

    Mass of the block (m) = 3.3 kg

    Spring constant of the spring (k) = 181 N/m

    Speed of the block at equilibrium length of the spring (v) = 5.5 m/s

    We know that, equilibrium length means there is no elongation or compression of the spring. So, there is no elastic potential energy and all the energy is due to kinetic energy of the block.

    Also, at maximum stretch of the spring, the kinetic becomes 0 and thus all the energy is due to elastic potential energy of the spring.

    Therefore, from conservation of energy, the decrease in kinetic energy is equal to increase in elastic potential energy.

    Kinetic energy is given as:

    K=\frac{1}{2}mv^2=\frac{1}{2}\times 3.3\times 5.5^2=49.912\ J

    So, elastic potential energy at maximum stretch is equal to 49.912 J.

    U=K\\\\\frac{1}{2}kx^2=49.912\\\\\frac{1}{2}\times 181\times x^2=49.912\\\\x^2=\frac{49.912\times 2}{181}\\\\x=\sqrt{\frac{99.824}{181}}\\\\x=0.742\approx0.74\ m(nearest\ hundredth)

    Therefore, the maximum amount that the spring can stretch is 0.74 m.

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