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Chapter 5: Problem 9

A particle moves parallel to the \(x\) -axis. The net force on the particle increases with \(x\) according to the formula \(F_{x}\) \(=(120 \mathrm{~N} / \mathrm{m}) x,\) where the force is in newtons when \(x\) is in meters. How much work does this force do on the particle as it moves from \(x=0\) to \(x=0.50 \mathrm{~m} ?\) a) 7.5 J c) \(30 J\) e) 120 J b) 15 J d) 60

Short Answer

Expert verified
Answer: The work done is 15 J.

Step by step solution

01

Recall the formula for work done by a variable force

The work done by a variable force \(F_{x}\) when moving from position \(x_{1}\) to \(x_{2}\) is given by: \[W = \int_{x_{1}}^{x_{2}} F_{x} dx\] In this case, the force function is \(F_{x} = 120x\), and we are asked to find the work done as the particle moves from \(x=0\) to \(x=0.50m\).
02

Set up the integral

Using the given force function and limits, we can set up the integral as follows: \[W = \int_{0}^{0.50} 120x dx\]
03

Evaluate the integral

Integrating the force function with respect to \(x\), we get: \[W = 60x^2\Big|_{0}^{0.50}\]
04

Compute the work

Now, apply the limits to find the work done: \[W = 60(0.50^2) - 60(0^2)\] \[W = 60(0.25)\] \[W = 15 J\] So, the correct answer is 15 J.

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