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Work done on box

[SOLVED] Work done on box

1. Homework Statement
An 8.00kg package in a mail-sorting room slides 2.00m down a chute that is inclined at 53.0 below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.400.

its asking me for the work done on the package by gravity and then the net work done on the package. wouldnt the work by gravity just be the weight. and for the normal force i just subtracted the frictional force from the normal force. where am i messing up?

2. Homework Equations
W= F cos(53) s
Friction Force= .4(Fnormal)

3. The Attempt at a Solution
I already figured out the frictional force to be 37.7 and the force 94J.
 
Last edited:

dynamicsolo

Homework Helper
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2. Homework Equations
W= F cos(53) s
Friction Force= .4(Fnormal)

3. The Attempt at a Solution
I already figured out the frictional force to be 37.7 and the force 94J.
I'm afraid this doesn't makes a lot of sense as it stands. Start with the forces:

You have the frictional force correct, but what is F_normal?

What is the weight force on the package? What component of that force points along the incline?

Of these three forces -- normal force, kinetic friction, and weight -- which ones will do work along the incline on the package?
 
For F_normal I have F_normal = (8kg)(9.8 m/s^2) cos53. is that correct?
and niether of those do work along the incline do they? the kinetic friction is going against it the weight is downwards and the F_normal is up right?
 

dynamicsolo

Homework Helper
1,648
4
For F_normal I have F_normal = (8kg)(9.8 m/s^2) cos53. is that correct?
That's right.

and niether of those do work along the incline do they? the kinetic friction is going against it the weight is downwards and the F_normal is up right?
The normal force is perpendicular to the direction the package moves, so it does not work on the package.

The kinetic friction force points up the slope and there is a component of the weight force that points down the incline. What are the magnitudes of those forces?
 
i thought the kinetic friction pointed opposite of the x direction? The component that points down would be the U=mgy correct?
 
ok i got that part and finished the rest of the problem thanks!
 

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