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Chapter 8: Problem 23

A stone used to grind wheat into flour is turned through 12 revolutions by a constant force of \(20.0 \mathrm{N}\) applied to the rim of a 10.0 -cm-radius shaft connected to the wheel. How much work is done on the stone during the 12 revolutions?

Short Answer

Expert verified
Answer: The work done on the stone during the 12 revolutions is approximately 150.8 J.

Step by step solution

01

Identify the given values

We are given the following values: - Force: \(F = 20.0 \mathrm{N}\) - Radius of the shaft: \(r = 10.0\ \mathrm{cm} = 0.1\ \mathrm{m}\) (converted to meters) - Number of revolutions: \(n = 12\)
02

Calculate the total distance traveled

First, we need to determine the distance the force is applied over the 12 revolutions. Since the force is applied at the rim of the shaft, the distance the force travels along the circular path can be found using the formula for the circumference of a circle, multiplied by the number of revolutions: Distance \(d = n \times C\) where \(C = 2 \pi r\) is the circumference of the circle. Substitute the given values: \(d = 12 \times (2 \pi \times 0.1)\) \(d = 12 \times (0.2 \pi)\) \(d \approx 7.54\ \mathrm{m}\)
03

Calculate the work done

Now that we have the distance, we can use the formula for work, which is given by: \(W = F \times d \times \cos \theta\) Since the force is applied tangent to the circle, it has an angle of \(90^{\circ}\) with the radial distance, and \(\cos 90^{\circ} = 0\). Therefore, the work done is: \(W = 20.0 \mathrm{N} \times 7.54\ \mathrm{m} \times 1\) \(W \approx 150.8\ \mathrm{J}\) So, the work done on the stone during the 12 revolutions is approximately \(150.8\ \mathrm{J}\).

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Most popular questions from this chapter

A weight of \(1200 \mathrm{N}\) rests on a lever at a point \(0.50 \mathrm{m}\) from a support. On the same side of the support, at a distance of $3.0 \mathrm{m}\( from it, an upward force with magnitude \)F$ is applied. Ignore the weight of the board itself. If the system is in equilibrium, what is \(F ?\)
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