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Chapter 5: Problem 22

A force of \(5.00 \mathrm{~N}\) acts through a distance of \(12.0 \mathrm{~m}\) in the direction of the force. Find the work done.

Short Answer

Expert verified
Answer: The work done by the force over the distance is 60.0 J.

Step by step solution

01

Write down the given values and formula

We are given: - Force, \(F = 5.00 \mathrm{~N}\) - Distance, \(d = 12.0 \mathrm{~m}\) - Angle, \(\theta = 0^\circ\) The formula for work done is: \(W = Fd\cos{\theta}\)
02

Calculate the cosine of the angle

As the angle is \(0^\circ\), the cosine of the angle is: \(\cos{0^\circ} = 1\)
03

Plug in the values into the formula and calculate the work done

Using the values and the cosine of the angle, the work done can be calculated as: \(W = (5.00 \mathrm{~N})(12.0 \mathrm{~m})(1)\) \(W = 60.0 \mathrm{~J}\) So, the work done by the force over the distance is \(60.0 \mathrm{~J}\).

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Most popular questions from this chapter

How much work is done against gravity in lifting a \(6.00-\mathrm{kg}\) weight through a distance of \(20.0 \mathrm{~cm} ?\)

Which of the following is a correct unit of power? a) \(\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}\) c) J e) \(W\) b) \(N\) d) \(\mathrm{m} / \mathrm{s}^{2}\)

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Which of the following is a correct unit of energy? a) \(\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}\) c) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}^{2}\) e) \(\mathrm{kg}^{2} \mathrm{~m}^{2} / \mathrm{s}^{2}\) b) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}\) d) \(\mathrm{kg}^{2} \mathrm{~m} / \mathrm{s}^{2}\)

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