About the concept of work done (High school student)

AI Thread Summary
The discussion centers on the concept of work done when pushing a box up a slope. The student is confused about whether the work done, calculated as force multiplied by distance, should include the gravitational potential energy gained by the box. It is clarified that the given force already accounts for both friction and the force needed to lift the box against gravity. To find the total energy transferred, one must consider all forces acting on the box, but the work done by the man is simply his force times the displacement. Understanding these distinctions is crucial for accurately calculating work and energy in physics.
FaroukYasser
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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?
 
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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.
 
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