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Chapter 5: Problem 38

An ideal spring has the spring constant \(k=440 \mathrm{~N} / \mathrm{m}\) Calculate the distance this spring must be stretched from its equilibrium position for 25 J of work to be done.

Short Answer

Expert verified
Answer: The spring must be stretched approximately 0.336 meters from its equilibrium position.

Step by step solution

01

Recall Hooke's Law

Hooke's law states that the force exerted by a spring is equal to the spring constant multiplied by the displacement from the equilibrium position: \(F = kx\) Where \(F\) is the force exerted by the spring, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. In this exercise, the spring constant \(k = 440 \mathrm{~N} / \mathrm{m}\).
02

Write the expression for the work done on the spring

The work done on the spring is given by the following equation: \(W = \frac{1}{2}kx^2\) Where \(W\) is the work done, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. In this problem, we are given that the work done, \(W\), is 25 J.
03

Substitute the known values and solve for the displacement x

We now have: \(25 \mathrm{J} = \frac{1}{2}(440 \mathrm{~N} / \mathrm{m})(x^2)\) First, we need to isolate \(x^2\) by dividing both sides of the equation by \(\frac{1}{2}k\): \(x^2 = \frac{50 \mathrm{J}}{440 \mathrm{~N} / \mathrm{m}}\) Next, we will calculate the fraction: \(x^2 = \frac{50}{440} \mathrm{m}\) Now, we can find the square root of both sides of the equation to solve for \(x\): \(x = \sqrt{\frac{50}{440}} \mathrm{m}\) Finally, let's approximate the distance: \(x \approx 0.336 \mathrm{m}\)
04

State the final answer

Therefore, the spring must be stretched approximately 0.336 meters from its equilibrium position for 25 J of work to be done.

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Most popular questions from this chapter

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