Need help on a work problem -- Pushing a shoe box up a ramp

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Homework Help Overview

The problem involves a constant force applied to a shoe box moving up a frictionless ramp at a constant speed. The scenario includes calculating the work done by the applied force when the box moves through a specified vertical distance.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the work done by the applied force and gravitational potential energy, questioning whether the work done is positive or negative.
  • Some participants explore the implications of constant speed on net work and kinetic energy, while others clarify the definitions of work done by gravity and the applied force.
  • There is confusion regarding the signs associated with gravitational force and work done, particularly in relation to the direction of forces.

Discussion Status

The discussion is ongoing, with participants seeking clarity on the signs of work done and the implications of the net work being zero. Some have provided insights into the definitions and relationships between the forces involved, but no consensus has been reached on the final interpretation of the work done.

Contextual Notes

Participants are navigating the complexities of sign conventions for forces and work, particularly in the context of a frictionless ramp and the effects of gravity. There is an acknowledgment of potential confusion stemming from the problem's setup and assumptions about initial conditions.

BlackJ
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Homework Statement


A constant force (Fa) of magnitude 82.0 N is applied to a 3.00 kg shoe box at an angle of 53.0 degrees, causing the box to move up a frictionless ramp at a constant speed. How much work is done on the box by Fa when the box has moved through a vertical distance h = 0.150 m.

Homework Equations

The Attempt at a Solution


My professor gave a hint that v is constant => a=0, thus, the Wnet= Wg + Wn + Wa =0 (Wn=0)
I feel confused here. If so, work done by Fa will be equal to negative potential energy Wg, which is -mgh and the answer will be in negative?
 
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BlackJ said:
negative potential energy Wg, which is -mgh
If work done is -Wg and Wg=-mgh, what is the work done?
 
Isn't it the work from gravity force?
 
BlackJ said:
Isn't it the work from gravity force?
I do not understand your question.
You are asked for the work done by Fa. Call this Wa.
There is an awkwardness in the question. If the box started from rest then it must have accelerated at some stage, so we have to assume it was already moving at speed v when we consider the force first applied. This means there is no change in KE and the net work done on the box is zero.
Gravity does work Wg on the box. Wa+Wg=0.
Wg=-mgh.
So what is Wa?
 
Sorry I thought you ask what Wg is.
I understand why the net work done on the box is zero, just still confused how the right answer is in positive but not negative after solving the equation Wa+Wg=0
As I see here
Wa+Wg= 0
=> Wa= -Wg= -mgh
Isn't the answer in negative?
 
BlackJ said:
Wa= -Wg= -mgh
No, Wg =-mgh, so what is -Wg?
 
does that mean g= -9.8m/s^2 instead of 9.8 because gravity applied a force in opposite direction of force a?
 
BlackJ said:
does that mean g= -9.8m/s^2 instead of 9.8 because gravity applied a force in opposite direction of force a?
I was taking g to be +9.8m/s2, since that is how most people use 'g'.
(If you are defining up as positive for everything else, it would be more logical to say the weight is mg where g=-9.8m/s2, but hardly anyone does that so let's go with the usual style.)
So, g=9.8m/s2, weight is -mg (because weight force acts down) and if an object rises by distance h the work done on it by gravity is -mgh.
Since the net work done on the mass in this question is zero, we have Wa+(-mgh)=0.
 
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Argh I kept thinking -Wg already included the negative sign so I substitute mgh instead of -mgh
It's just clear now, thanks a lot.
 

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