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Chapter 6: Problem 62

A person is making homemade ice cream. She exerts a force of magnitude \(22 \mathrm{N}\) on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius \(0.28 \mathrm{m}\). The force is always applied parallel to the motion of the handle. If the handle is turned once every \(1.3 \mathrm{s},\) what is the average power being expended?

Short Answer

Expert verified
The average power expended is approximately 29.71 W.

Step by step solution

01

Calculate the Work Done per Revolution

Work is done when a force moves an object along a path. Here, the force is applied in a circular path. The formula for work done when a force is applied along a circular path is:\[W = F \times 2\pi r\]where \( W \) is the work done, \( F = 22 \text{ N} \) is the force, and \( r = 0.28 \text{ m} \) is the radius of the circular path. Substituting the known quantities:\[W = 22 \times 2 \pi \times 0.28 = 38.624 \text{ J}\]
02

Determine the Time for One Revolution

The problem states that the handle is turned once every 1.3 seconds. Thus, the time for one complete revolution is \( t = 1.3 \text{ s} \).
03

Calculate the Average Power

Power is defined as the rate at which work is done. The formula for average power is:\[P = \frac{W}{t}\]Using \( W = 38.624 \text{ J} \) and \( t = 1.3 \text{ s} \):\[P = \frac{38.624}{1.3} \approx 29.71 \text{ W}\]
04

Interpret the Result

This means that the average power being expended as the crank handle rotates is approximately \( 29.71 \text{ W} \). Power is the rate of energy conversion, and in this case, indicates how quickly the person is doing work on the ice cream maker.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Done in Circular Path
When it comes to understanding work done in a circular path, it's important to realize that work is only done when a force causes displacement. In this scenario, a force is applied along a circular route as a person turns a crank. This setup is quite common in real-world applications, like stirring or turning tools and gadgets.

The amount of work done is remarkably calculated using the formula:
  • Work ( \(W\)) = Force ( \(F\)) \( \times \) Circumference of the Circle
  • For a circle, the circumference is \(2 \pi r\), where \(r\) is the radius.
This means our total work formula becomes:
  • \(W = F \times 2 \pi r\)
Applying this formula allows you to determine how much energy is transferred or converted in each cycle. Knowing this is key to understanding many physical processes, especially when dealing with machines that require repetitive motion.
Average Power
Power is all about the rate at which work is performed or energy is converted. When dealing with circular motion and repeated cycles like turning a crank handle, it becomes crucial to calculate average power.

The average power, \(P\), can be expressed as:
  • Average Power ( \(P\)) = Work Done ( \(W\)) / Time Taken ( \(t\))
In this context, power helps us understand how intensely energy is being used or transferred. For our exercise:
  • Total work per revolution is \(38.624 \,J\)
  • The time for one cycle is \(1.3 \,s\)
  • Therefore, average power is \( \approx 29.71 \,W\)
Calculating average power is useful, because it provides a measure of efficiency and performance over repeated tasks. It's a foundational concept in physics that applies to everything from simple machines to advanced technologies.
Force and Motion
In physics, understanding the relationship between force and motion is vital. Force is what causes objects to start moving, stop, or change direction. When force acts along a motion path, as in our circular motion example, it can result in work being done.

For circular motion:
  • The force applied must be tangent to the path. This ensures movement along the circular route.
  • In our case, the force applied is constantly parallel to the trajectory of the handle.
This alignment of force and motion ensures that energy is efficiently transferred to perform the desired action. Understanding how they interact helps us design tools and machines that operate effectively under different conditions.
Energy Conversion
Energy conversion is a key theme in many physics problems. It's the process of changing energy from one form to another. When making ice cream, mechanical energy is transferred into the crank, bringing about motion and performing work on the ice cream mixture.

In this scenario, the energy conversion process is clear:
  • Mechanical energy from the arm results in the rotation of the crank.
  • This mechanical energy is used to churn the ice cream, converting it to potential energy within the mixture.
These transformations reflect fundamental principles of conservation of energy, ensuring energy isn't lost but redirected and re-applied in different forms. This type of conversion is central to how many mechanical devices operate, pointing to underlying principles that power technology and engines globally.

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Most popular questions from this chapter

The drawing shows two frictionless inclines that begin at ground level \((h=0 \mathrm{m})\) and slope upward at the same angle \(\theta\). One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed \(v_{0} .\) On the longer track the block slides upward until it reaches a maximum height \(H\) above the ground. On the shorter track the block slides upward, flies off the end of the track at a height \(H_{1}\) above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height \(H_{2}\) above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. The initial speed of each block is \(v_{0}=7.00 \mathrm{m} / \mathrm{s},\) and each incline slopes upward at an angle of \(\theta=50.0^{\circ} .\) The block on the shorter track leaves the track at a height of \(H_{1}=1.25 \mathrm{m}\) above the ground. Find (a) the height \(H\) for the block on the longer track and (b) the total height \(H_{1}+H_{2}\) for the block on the shorter track.

It takes \(185 \mathrm{kJ}\) of work to accelerate a car from \(23.0 \mathrm{m} / \mathrm{s}\) to \(28.0 \mathrm{m} / \mathrm{s} .\) What is the car's mass?

A truck is traveling at \(11.1 \mathrm{m} / \mathrm{s}\) down a hill when the brakes on all four wheels lock. The hill makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is \(0.750 .\) How far does the truck skid before coming to a stop?

At a carnival, you can try to ring a bell by striking a target with a 9.00-kg hammer. In response, a \(0.400-\mathrm{kg}\) metal piece is sent upward toward the bell, which is \(5.00 \mathrm{m}\) above. Suppose that \(25.0 \%\) of the hammer's kinetic energy is used to do the work of sending the metal piece upward. How fast must the hammer be moving when it strikes the target so that the bell just barely rings?

A projectile of mass 0.750 kg is shot straight up with an initial speed of \(18.0 \mathrm{m} / \mathrm{s}\) (a) How high would it go if there were no air resistance? (b) If the projectile rises to a maximum height of only \(11.8 \mathrm{m}\), determine the magnitude of the average force due to air resistance.
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