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Chapter 13: Problem 4

Weiping lifts a rock with a weight of \(1.0 \mathrm{~N}\) through a height of \(1.0 \mathrm{~m}\), and then lowers it back down to the starting point. Bubba pushes a table \(1.0 \mathrm{~m}\) across the floor at constant speed, requiring a force of \(1.0 \mathrm{~N}\), and then pushes it back to where it started. (a) Compare the total work done by Weiping and Bubba. (b) Check that your answers to part a make sense, using the definition of work: work is the transfer of energy. In your answer, you'll need to discuss what specific type of energy is involved in each case.

Short Answer

Expert verified
Weiping and Bubba both do 0 J of total work, as the work cancels out in each scenario.

Step by step solution

01

Calculate Work Done by Weiping

Work is calculated using the equation \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force, \( d \) is the distance, and \( \theta \) is the angle between the force and the direction of motion. For Weiping lifting the rock, \( F = 1.0 \ \mathrm{N} \), \( d = 1.0 \ \mathrm{m} \), and \( \theta = 0 \) degrees, as the force is in the same direction as the motion. Thus, the work done in lifting the rock is \( W_{up} = 1.0 \ \mathrm{N} \times 1.0 \ \mathrm{m} = 1.0 \ \mathrm{J} \). When lowering the rock, the force is also 1.0 N, and the motion is in the same direction, so \( W_{down} = 1.0 \ \mathrm{J} \), but negative since the work is in the opposite direction to the displacement. The total work done is \( 1.0 \ \mathrm{J} - 1.0 \ \mathrm{J} = 0 \ \mathrm{J} \).
02

Calculate Work Done by Bubba

Similar to Weiping's scenario, the work done by Bubba when pushing the table forward is \( W_{forward} = 1.0 \ \mathrm{N} \times 1.0 \ \mathrm{m} = 1.0 \ \mathrm{J} \). When Bubba pushes the table back to the starting point, he does \( W_{back} = 1.0 \ \mathrm{J} \), but this is also in the positive direction and cancels out with the first movement. Thus, the total work done by Bubba is \( 1.0 \ \mathrm{J} + 1.0 \ \mathrm{J} = 0 \ \mathrm{J} \).
03

Compare Total Work

Weiping's total work is 0 J, and Bubba's total work is also 0 J. Therefore, the total work done by Weiping and Bubba is the same, both being 0 J.
04

Discuss Energy Transfer

Work is defined as the transfer of energy. For Weiping, the energy transferred when lifting the rock is potential energy (gravitational), which is then converted back into kinetic energy as the rock is lowered again. For Bubba, since the table is moved at constant speed, the energy transferred is kinetic energy, which is used to overcome friction, and the back and forth motion results in no net energy transfer over the entire cycle.

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Key Concepts

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

Potential Energy
Potential energy is an essential concept in understanding how objects store energy due to their position or arrangement. In the case of Weiping lifting a rock, when the rock is raised to a height of 1 meter, it gains potential energy due to the gravitational force acting on it. This energy depends on three main factors:

  • The mass of the object
  • The gravitational force
  • The height above the ground
Since the weight of the rock is given as 1.0 N and it is lifted 1.0 meter upward, the potential energy (9) gained by the rock is equivalent to the work done against gravity, which is 1.0 Joule.

However, when Weiping lowers the rock back to its original position, this potential energy is converted back, showing how energy can be stored and released through position changes. This concept is crucial in analyzing energy conservation and transfer in physics, particularly when dealing with gravitational fields.
Kinetic Energy
Kinetic energy is the energy of motion. It is the energy an object possesses because of its movement. For Bubba, when he pushes the table across the floor at a constant speed, he applies a force to overcome friction, thereby transferring energy to keep the table moving.

The kinetic energy can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
  • \( m \) is the mass of the object
  • \( v \) is the velocity of the object
In Bubba's case, the exact velocity isn't provided, but continuous application of force is required to maintain motion against friction鈥攖his indicates energy is consistently being used, which is a testament to kinetic energy's role.

Understanding kinetic energy helps clarify why continuous energy input is needed to maintain constant velocity, especially when countering frictional forces.
Energy Transfer
Energy transfer is a fundamental process where energy changes from one form to another. In physics, it plays a crucial role in understanding how work and energy are related.

When Weiping lifts the rock, energy is transferred from his muscles to the rock, converting from chemical energy in his body into gravitational potential energy. As the rock is lowered, this potential energy is released and converted back into kinetic energy as it moves downwards.

For Bubba, the energy transfer is slightly different. He uses energy to push the table, transferring stored energy from his muscles to kinetic energy, enough to overcome friction and move the table. Though the net work done is zero because the table returns to its original position, energy is still transferred by doing work against friction, effectively showing how energy transfer doesn't always result in a permanent change in energy state or location.

This demonstrates the principles of conservation and conversion of energy, which are key in understanding physical movements and mechanics of everyday activities.

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Most popular questions from this chapter

A car accelerates from rest. At low speeds, its acceleration is limited by static friction, so that if we press too hard on the gas, we will "burn rubber" (or, for many newer cars, a computerized traction-control system will override the gas pedal). At higher speeds, the limit on acceleration comes from the power of the engine, which puts a limit on how fast kinetic energy can be developed. (a) Show that if a force \(F\) is applied to an object moving at speed \(v\), the power required is given by \(P=v F\). (b) Find the speed \(v\) at which we cross over from the first regime described above to the second. At speeds higher than this, the engine does not have enough power to burn rubber. Express your result in terms of the car's power \(P\), its mass \(m\), the coefficient of static friction \(\mu_{s}\), and \(g\). (c) Show that your answer to part b has units that make sense. (d) Show that the dependence of your answer on each of the four variables makes sense physically. (e) The 2010 Maserati Gran Turismo Convertible has a maximum power of \(3.23 \times 10^{5} \mathrm{~W}\) (433 horsepower) and a mass (including a 50\(\mathrm{kg}\) driver ) of \(2.03 \times 10^{3} \mathrm{~kg}\). (This power is the maximum the engine can supply at its optimum frequency of 7600 r.p.m. Presumably the automatic transmission is designed so a gear is available in which the engine will be running at very nearly this frequency when the car is moving at \(v .\) ) Rubber on asphalt has \(\mu_{s} \approx 0.9\). Find \(v\) for this car. Answer: \(18 \mathrm{~m} / \mathrm{s}\), or about 40 miles per hour. (f) Our analysis has neglected air friction, which can probably be approximated as a force proportional to \(v^{2} .\) The existence of this force is the reason that the car has a maximum speed, which is 176 miles per hour. To get a feeling for how good an approximation it is to ignore air friction, find what fraction of the engine's maximum power is being used to overcome air resistance when the car is moving at the speed \(v\) found in part e. Answer: \(1 \%\)

(a) The crew of an 18th century warship is raising the anchor. The anchor has a mass of \(5000 \mathrm{~kg}\). The water is \(30 \mathrm{~m}\) deep. The chain to which the anchor is attached has a mass per unit length of \(150 \mathrm{~kg} / \mathrm{m}\). Before they start raising the anchor, what is the total weight of the anchor plus the portion of the chain hanging out of the ship? (Assume that the buoyancy of the anchor is negligible.) (b) After they have raised the anchor by \(1 \mathrm{~m}\), what is the weight they are raising? (c) Define \(y=0\) when the anchor is resting on the bottom, and \(y=+30 \mathrm{~m}\) when it has been raised up to the ship. Draw a graph of the force the crew has to exert to raise the anchor and chain, as a function of \(y\). (Assume that they are raising it slowly, so water resistance is negligible.) It will not be a constant! Now find the area under the graph, and determine the work done by the crew in raising the anchor, in joules. (d) Convert your answer from (c) into units of kcal.

A soccer ball of mass \(m\) is moving at speed \(v\) when you kick it in the same direction it is moving. You kick it with constant force \(F\), and you want to triple the ball's speed. Over what distance must your foot be in contact with the ball?

In 1935 , Yukawa proposed an early theory of the force that held the neutrons and protons together in the nucleus. His equation for the potential energy of two such particles, at a center-tocenter distance \(r\), was \(P E(r)=g r^{-1} e^{-r / a}\), where \(g\) parametrizes the strength of the interaction, \(e\) is the base of natural logarithms, and \(a\) is about \(10^{-15} \mathrm{~m}\). Find the force between two nucleons that would be consistent with this equation for the potential energy.

(a) Suppose work is done in one-dimensional motion. What happens to the work if you reverse the direction of the positive coordinate axis? Base your answer directly on the definition of work as a transfer of mechanical energy. (b) Now answer the question based on the \(W=F d\) rule.
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