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Chapter 5: Problem 34

A particle of mass \(m\) is subjected to a force acting in the \(x\) -direction. \(F_{x}=(3.0+0.50 x) \mathrm{N}\). Find the work done by the force as the particle moves from \(x=0\) to \(x=4.0 \mathrm{~m}\)

Short Answer

Expert verified
Answer: The work done by the force is \(16 \, Joules\).

Step by step solution

01

1. Set up the integral expression for the work done.

To find the work done by the force, write down the integral expression for the work done over the interval \([0, 4]\): \(W = \int_{0}^{4} F(x) dx = \int_{0}^{4} (3.0 + 0.50x) dx\).
02

2. Evaluate the integral.

Now, evaluate the integral by finding the antiderivative of the force function and evaluating at the bounds. To find the antiderivative of \(3.0 + 0.50x\), we have: \(\int (3.0 + 0.50x) dx = 3.0x + 0.25x^2 + C\), where \(C\) is the constant of integration. As we are evaluating this between definite limits, the constant \(C\) has no effect on our calculation. Now, evaluate at the bounds: \(W = [3.0x + 0.25x^2]_0^4 = (3.0(4) + 0.25(4^2)) - (3.0(0) + 0.25(0^2))\).
03

3. Calculate the work done.

Calculate the work done by simplifying the expression: \(W = (12 + 0.25(16)) - (0 + 0) = 12 + 4 = 16 J\). So the work done by the force as the particle moves from \(x = 0\) to \(x = 4.0 m\) is \(16 \, Joules\).

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Most popular questions from this chapter

A spring with spring constant \(k\) is initially compressed a distance \(x_{0}\) from its equilibrium length. After returning to its equilibrium position, the spring is then stretched a distance \(x_{0}\) from that position. What is the ratio of the work that needs to be done on the spring in the stretching to the work done in the compressing?

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