Work done on package by gravity: 784JNet work done on package: -37.7J

In summary: The normal force is perpendicular to the direction the package moves, so it does not work on the package.
  • #1
bigtymer8700
40
0
[SOLVED] Work done on box

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. wouldn't 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?

Homework Equations


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

The Attempt at a Solution


I already figured out the frictional force to be 37.7 and the force 94J.
 
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  • #2
bigtymer8700 said:

Homework Equations


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

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?
 
  • #3
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?
 
  • #4
bigtymer8700 said:
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?
 
  • #5
i thought the kinetic friction pointed opposite of the x direction? The component that points down would be the U=mgy correct?
 
  • #6
ok i got that part and finished the rest of the problem thanks!
 

1. What is work done on a box?

Work done on a box is the amount of energy that is transferred to the box when a force is applied to move it a certain distance.

2. How is work calculated on a box?

The formula for calculating work on a box is work = force x distance. The force should be in the same direction as the displacement of the box.

3. Does the weight of the box affect the work done on it?

Yes, the weight of the box is a force acting on it, so it will affect the work done on it. The work done will be greater if the box is heavier.

4. Is work done on a box always positive?

No, work can be either positive or negative depending on the direction of the applied force. If the force is in the same direction as the displacement, work will be positive. If the force is in the opposite direction, work will be negative.

5. Can work be done on a box if it is not moving?

Yes, work can still be done on a box even if it is not moving. This is because work is a measure of energy transfer, not necessarily movement. For example, if a person pushes a box but it does not move, work is still being done on the box as energy is being transferred from the person's muscles to the box.

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