Work Up an Incline: Calculate Gravity, Friction & Push Work

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Homework Help Overview

The problem involves calculating the work done by gravity, friction, and a person pushing a box up an incline. The box has a mass of 50 kg, is pushed up a 25-degree incline, and reaches a height of 45 m, with a coefficient of kinetic friction of 0.4.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the appropriate distance to use for calculating work, questioning whether to use the vertical height or the distance along the incline. There is also consideration of how to calculate work done by gravity in relation to height versus distance on the incline.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the relevant equations. Some guidance has been offered regarding potential approaches to calculating work done by gravity and friction, but no consensus has been reached.

Contextual Notes

Participants are navigating the complexities of the problem, including the definitions of distance in the context of work and the implications of the incline's angle on calculations.

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Homework Statement


A 50 kg box is being pushed up an incline of 25 degrees, to a total height of 45 m. Kinetic friction between the box and the incline has a mew of .4.
a) What is the work done by gravity?
b)What is the work done by friction?
c) What is the work done by the person pushing the box.



Homework Equations


W = Fd


The Attempt at a Solution


For all of the parts, I am not sure what to use as my d. Do I use 45, or do i use (45/sin(45)), the hypotenuse and the distance that the box actually slides?
 
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keemosabi said:

Homework Statement


A 50 kg box is being pushed up an incline of 25 degrees, to a total height of 45 m. Kinetic friction between the box and the incline has a mew of .4.
a) What is the work done by gravity?
b)What is the work done by friction?
c) What is the work done by the person pushing the box.

Homework Equations


W = Fd

The Attempt at a Solution


For all of the parts, I am not sure what to use as my d. Do I use 45, or do i use (45/sin(45)), the hypotenuse and the distance that the box actually slides?

For changes in PE it's straight height difference.

For friction it's distance along the incline.
 
LowlyPion said:
For changes in PE it's straight height difference.

For friction it's distance along the incline.
What about the work done by gravity? Is that height or distance on the incline? Or does it not matter.
 
keemosabi said:
What about the work done by gravity? Is that height or distance on the incline? Or does it not matter.

Work by gravity is change in PE = m*g*h
 

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