Zero work illustrative problem

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In summary, the conversation discusses the concept of work being done on an object by a force. A boy is holding a book and then starts walking with it, causing it to move horizontally. The question is whether there is work being exerted by the boy on the book. The answer is that there is a small amount of work done to initially start the book moving, but once it is moving at a constant velocity, there is no work being done. However, there may be other forces acting on the book, such as gravity and friction, which may result in a net force of 0 N. In this case, the work done by these forces would also be 0 J. The conversation also discusses a scenario where a force is applied
  • #1
arjose
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Homework Statement



A boy supports a book on his hand while standing still. Then, he walks with the book laying motionless on his hand. Is there work exerted by the boy on the book?

2. The attempt at a solution

My professor said that there is zero work done on the book because of the perpendicularity of the displacement of the book and the force applied by the boy on the book. But, I ask, if the book moved from rest, doesn't it mean it accelerated? If it accelerated, doesn't it mean that a net force along the x-axis acted on it? If so, then doesn't it mean that there IS work? Please help.
 
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  • #2
arjose said:

Homework Statement



A boy supports a book on his hand while standing still. Then, he walks with the book [strike]laying[/strike] lying motionless on his hand. Is there work exerted by the boy on the book?

2. The attempt at a solution

My professor said that there is zero work done on the book because of the perpendicularity of the displacement of the book and the force applied by the boy on the book. But, I ask, if the book moved from rest, doesn't it mean it accelerated? If it accelerated, doesn't it mean that a net force along the x-axis acted on it? If so, then doesn't it mean that there IS work? Please help.
Yes, you're right. There was work done on the book to get it to move horizontally.

Once the book is moving, if it's moving at constant velocity, there's no work being done on it by the boy because the force exerted by his hand is perpendicular to the displacement.
 
  • #3
vela said:
Once the book is moving, if it's moving at constant velocity, there's no work being done on it by the boy because the force exerted by his hand is perpendicular to the displacement.

Given that the book is moving at constant velocity, does it matter if the force exerted by his hand is counteracted by the weight of the book? I mean is there even work to be computed if net vertical force on the book is zero?
 
  • #4
arjose said:
Given that the book is moving at constant velocity, does it matter if the force exerted by his hand is counteracted by the weight of the book? I mean is there even work to be computed if net vertical force on the book is zero?

The net vertical force is zero. What do you think about the horizontal direction?
Which forces are acting on the book then (2 forces)?

When he starts walking there is acceleration, and some work is involved.

Is there work done thereafter? The answer depends on the sizes of 2 forces in the horizontal direction, which determine whether he then walks with constant acceleration (sum of acting forces zero) or has acceleration (sum of acting forces other than zero)
 
  • #5
arjose said:

Homework Statement



A boy supports a book on his hand while standing still. Then, he walks with the book laying motionless on his hand. Is there work exerted by the boy on the book?

2. The attempt at a solution

My professor said that there is zero work done on the book because of the perpendicularity of the displacement of the book and the force applied by the boy on the book. But, I ask, if the book moved from rest, doesn't it mean it accelerated? If it accelerated, doesn't it mean that a net force along the x-axis acted on it? If so, then doesn't it mean that there IS work? Please help.

Strictly speaking there is a little bit of work done on the book starting it moving when the boy walks off, then a bit more stopping it again when the boy stops.
What you professor is referring to is that the Force supporting the book, perhaps 10N if the book has a mass of 1 kg, does no work as the boy carried the book 5m across a room. Some people would be inclined to think that since work if force x distance, then 50 J of work was done in the above situation. That reasoning is incorrect.
 
  • #6
arjose said:
Given that the book is moving at constant velocity, does it matter if the force exerted by his hand is counteracted by the weight of the book? I mean is there even work to be computed if net vertical force on the book is zero?
Yes, you can calculate the work done by the force the boy exerts and the work done by gravity separately. In this case, they both turn out to be 0 J so it's not very interesting.

You will likely do a problem where, say, a force is exerted on an object, causing it to move across the ground with constant velocity. The net force on the object is 0 N, but you'll be asked to calculate the work done by the applied force and the work done by friction separately. You'll find that neither is 0 J, but they cancel so that the total work done is 0 J.
 
  • #7
vela said:
You will likely do a problem where, say, a force is exerted on an object, causing it to move across the ground with constant velocity. The net force on the object is 0 N, but you'll be asked to calculate the work done by the applied force and the work done by friction separately. You'll find that neither is 0 J, but they cancel so that the total work done is 0 J.

Those underlined got me confused again. How is the object moving at just a constant velocity if net forces cause acceleration? I'm thinking a force moved it from rest. If so, then there was acceleration, which also means a nonuniform velocity.
 
  • #8
I didn't say there was a net force exerted on the object, just a force, like someone pushing a box across the floor. Other forces, like friction from the floor, act on the object as well, which result in a net force of 0 N.
 
  • #9
vela said:
I didn't say there was a net force exerted on the object, just a force, like someone pushing a box across the floor. Other forces, like friction from the floor, act on the object as well, which result in a net force of 0 N.

Ah. Now i get it. Thanks, vela.
 

1. What is a "zero work illustrative problem"?

A zero work illustrative problem is a type of physics problem that involves calculating the amount of work done on an object when the net force acting on the object is equal to zero. This means that there is no change in the object's kinetic energy, and thus no work is being done on the object.

2. How do you solve a zero work illustrative problem?

To solve a zero work illustrative problem, you will need to use the formula W = Fd, where W is the work done, F is the net force, and d is the distance the object moves. You will also need to determine the direction of the net force and whether it is positive or negative, as this will affect the final answer.

3. What is the significance of solving a zero work illustrative problem?

Solving a zero work illustrative problem allows you to understand the concept of work and how it is affected by different forces. It also helps you develop problem-solving skills and apply theoretical knowledge to real-world situations.

4. Can you provide an example of a zero work illustrative problem?

Sure! Imagine a car driving on a flat road at a constant speed. In this scenario, the net force acting on the car is equal to zero, as the opposing forces of friction and air resistance cancel each other out. Therefore, the work done on the car is also equal to zero.

5. How does a zero work illustrative problem relate to Newton's Second Law of Motion?

A zero work illustrative problem is directly related to Newton's Second Law of Motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In a zero work problem, the net force is equal to zero, meaning the acceleration is also equal to zero, and there is no change in the object's velocity or kinetic energy.

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