A 46.5\ \mathrm{kg} block sits at rest on a frictionless, horizontal surface. It is connected...

Question:

A {eq}46.5\ \mathrm{kg} {/eq} block sits at rest on a frictionless, horizontal surface. It is connected horizontally, via a massless, ideal spring to a stationary wall. A force is then used to displace and hold the block in a new position (so that the spring is compressed). Then the block is released to move freely. In that first instant of motion, its acceleration is {eq}7.98\ \mathrm{m/s^2} {/eq}, and when it passes through its original (first) position, its speed is {eq}13.2\ \mathrm{m/s} {/eq}. What is the springs constant, {eq}k {/eq}?

Spring-Mass System:

A spring-mass arrangement analysis is done on the principle of simple harmonic motion. The movement and the oscillation can be obtained from this system using simple harmonic analysis. The acceleration of this system establishes the relation with the amplitude of the system.

Answer and Explanation:

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Given data

  • The mass of the block of the system is: {eq}m = 46.5\;{\rm{kg}} {/eq}
  • The acceleration of the system is: {eq}a = 7.98\;{{\rm{m}}...

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Hooke's Law & the Spring Constant: Definition & Equation

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Chapter 4 / Lesson 19
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After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.


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