About the concept of work done (High school student)

Click For Summary

Discussion Overview

The discussion revolves around the concept of work done in the context of a man pushing a box up a slope. Participants explore the relationship between the applied force, the distance moved, and the energy changes involved, particularly regarding gravitational potential energy and friction. The scope includes conceptual clarification and mathematical reasoning related to work and energy in physics.

Discussion Character

  • Conceptual clarification
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about whether the work done formula (Force x Distance) accounts for changes in gravitational potential energy when pushing the box up the slope.
  • Another participant suggests that the given force includes both the force needed to overcome friction and the force required to lift the box against gravity.
  • A later reply proposes that to compute total energy transferred, one should consider the components of each force and their respective displacements, questioning if the formula should be interpreted as resultant force by displacement.
  • It is noted that to determine the work done by the man, one can use the force he exerts times the displacement, but calculating the final energy of the box requires considering all acting forces, including gravity and friction.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the work done formula inherently includes all energy changes. There are multiple competing views regarding the interpretation of the forces involved and how they relate to the work done and energy transfer.

Contextual Notes

Participants discuss the implications of different forces acting on the box, including friction and gravity, but do not resolve how these forces quantitatively affect the work done calculation.

FaroukYasser
Messages
62
Reaction score
3
Hi I am kind of Confused about an example in work done:

It says that a man applies 250 N parallel to the slope to push a box up a slope for 3 m, Now I know that the formula for work done if Force by Perpendicular distance which is 250 x 3 = 750 (which is the correct answer in my book). The thing is, isn't the object "Gaining" Gravitational potential energy ie change in its energy ie transfer of energy and therefor we should add that energy to the 750??
This is where I am confused.
Can anyone help please :)Edit: Or is the Force x Distance formula includes all changes in energy together?
 
Physics news on Phys.org
FaroukYasser said:
Hi I am kind of Confused about an example in work done:

It says that a man applies 250 N parallel to the slope to push a box up a slope for 3 m, Now I know that the formula for work done if Force by Perpendicular distance which is 250 x 3 = 750 (which is the correct answer in my book). The thing is, isn't the object "Gaining" Gravitational potential energy ie change in its energy ie transfer of energy and therefor we should add that energy to the 750??
This is where I am confused.
Can anyone help please :)


Edit: Or is the Force x Distance formula includes all changes in energy together?

Welcome to the PF.

Since you are explicitly given the force needed to push the box up the ramp, that includes both the force needed to push against the friction of the box on the ramp, and the force needed to lift the box up the delta-height that it gains. Does that make more sense?
 
berkeman said:
Welcome to the PF.

Since you are explicitly given the force needed to push the box up the ramp, that includes both the force needed to push against the friction of the box on the ramp, and the force needed to lift the box up the delta-height that it gains. Does that make more sense?

So what you mean is that if we get the component of each force and see the distance moved by each to get the total energy transferred it would add up to the Force x distance? so this formula should really be
Resultant force by displacement right?
 
FaroukYasser said:
So what you mean is that if we get the component of each force and see the distance moved by each to get the total energy transferred it would add up to the Force x distance? so this formula should really be
Resultant force by displacement right?
It depends on what you want to compute.

If you want the work done by the man, then that is simply the force he exerts times the displacement. But to determine the final energy of the box, you'll need to consider all the forces acting, which includes gravity and friction.
 

Similar threads

  • · Replies 48 ·
2
Replies
48
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 55 ·
2
Replies
55
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
5K