Calculating Work: When to Use W = Fd cos(a) vs. the Integral of F

  • Thread starter Thread starter caspeerrr
  • Start date Start date
  • Tags Tags
    Work
Click For Summary
SUMMARY

The discussion clarifies the application of the work formula W = Fd cos(a) versus the integral of force (F) in physics. W = Fd cos(a) is applicable when the force is constant and acts along a straight path, such as pushing an object at a constant speed. Conversely, the integral of F is used when the force varies along the path of integration, such as in scenarios involving changing forces. Examples provided illustrate these distinctions effectively.

PREREQUISITES
  • Understanding of basic physics concepts, particularly work and force.
  • Familiarity with the formula W = Fd cos(a).
  • Knowledge of integration in calculus.
  • Ability to analyze problems involving variable forces.
NEXT STEPS
  • Study examples of constant force scenarios using W = Fd cos(a).
  • Explore problems involving variable forces and the application of the integral of F.
  • Review calculus techniques for integrating force functions.
  • Examine real-world applications of work calculations in physics.
USEFUL FOR

Students in physics courses, educators teaching mechanics, and anyone seeking to understand the principles of work and force in physical systems.

caspeerrr
Messages
9
Reaction score
0

Homework Statement


When do you use W = Fd cos(a) and when do you use the integral of F?

Homework Equations


W = Fd cos(a) Integral of F

The Attempt at a Solution


When doing exercises it seems random to me when I have to use the formule W = fd cos(a) and when I have to use the integral? What is the consistency in this?
Thanks!
 
Physics news on Phys.org
caspeerrr said:

Homework Statement


When do you use W = Fd cos(a) and when do you use the integral of F?

Homework Equations


W = Fd cos(a) Integral of F

The Attempt at a Solution


When doing exercises it seems random to me when I have to use the formule W = fd cos(a) and when I have to use the integral? What is the consistency in this?
Thanks!
It would help if you could give a few examples. Can you post 1-2 of each type of problem? In general you integrate if something is changing along the path of integration. If things are constant along the path (like pushing a block at a constant speed across a floor), then you can just use W=Fd.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
9
Views
3K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
11K
Replies
18
Views
4K