No Work Done on a Textbook: Explanation of Concept

  • Thread starter Thread starter Hockey101
  • Start date Start date
  • Tags Tags
    Concept Work
AI Thread Summary
When carrying a textbook horizontally at a constant velocity, no work is done because the force applied (gravity) acts vertically while the displacement is horizontal. Work is defined as the product of force and displacement in the direction of the force, represented by the formula W = FΔs cos θ. Since the angle between the vertical force and horizontal displacement is 90 degrees, the cosine of 90 degrees is zero, resulting in no work being performed. If a horizontal force were applied in the same direction as the displacement, work would be done on the textbook. Understanding this relationship between force and displacement is crucial for grasping the concept of work in physics.
Hockey101
Messages
8
Reaction score
0
1. Hey everyone! I am currently studying for my final exam and I am trying to review work to start out. When I was reading, it gave me an example of where a person is holding a textbook and the person moves the textbook in the horizontal direction. Under the picture they gave an explanation which says: No work is done on a textbook when carried in the horizontal direction at a constant velocity.



2. None



3. Considering Work depends on displacement, I just thought that if you move a textbook up or down and that is displacement, then why can't there be displacement if you move the textbook (or anything) horizontally?
 
Physics news on Phys.org
Do you know the definition of "work"? I mean, do you know how to calculate it? What do you get if you use that formula to calculate the work performed moving a book horizontally?
 
From the site that I am reading from...Work is the "force-displacement product". The formula shows to be: W = FΔs cos θ
 
Ok...so is it because since there is a vertical force and the displacement is horizontal, then that's why work is not being done? So with that in mind, if there was a horizontal force and a horizontal displacement, then there would be work being done on the textbook?
 
Right. The ##\theta## in that formula is the angle between the force vector and the displacement vector. So when the force and the displacement are perpendicular, we have ##\cos\theta=\cos\frac\pi 2=0##. When they are parallel, we have ##\cos\theta=\cos 0=1##.
 
  • Like
Likes 1 person
Thank you so much!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top