What is the net work done on the box?

In summary, a man picks up a 10kg box and carries it from A to F, with a height difference of 3m, without doing any work due to the perpendicular force of carrying it. The net work done on the box is 0. However, in reality, it is not possible to displace the box without doing work due to friction.
  • #1
JWSiow
21
0

Homework Statement


A man picks up a 10kg box, and holds it at a height of 1m above the horizontal ground as he walks to a ramp, and then carries the box up the ramp to a height of 4m, and then down the ramp as shown, finally placing it on the ground at F.
21c72jd.jpg

Describe the work done at each stage of the motion. What is the net work done on the box?

Homework Equations


W=Fdcos[tex]\theta[/tex]


The Attempt at a Solution


Not really sure with this one.

I said A-->B = no work done. B-->C = work done (don't think that's a good description though!). C-->D = no work done. D-->E = work done (would this be negative (or positive) work? E-->F = no work done.

So, the net work done on the box would = 0.
 
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  • #2


Do you mean to say that the box is not displaced at all (i.e., is it lying at the same position at the end)?
Obviously no!
 
  • #3


I don't know, I don't really get the question. In the textbook, it says that you carry something in your hands, you don't do any work on it since the force to carry it is perpendicular to the displacement.
 
  • #4


Anyone? :S I need to know by tomorrow! (have an exam...)
 
  • #5


But in the textbook, it said, "You also do no work on the bag of groceries if you carry it as you walk horizontally across the floor at constant velocity"... which is why I thought no work would be done from A to B, C to D and E to F.

How would I work out the net work done on the box?
 
  • #6


Yeah, your textbook is correct. Even though the box is displaced from A to F, the force is at right angles to direction of displacement.
So the work done is zero as you mentioned.
 
  • #7


Ok, thanks! That makes me feel much more relaxed! (cos it seemed like something that could come up in the exam) :)
 
  • #8


n.karthick said:
Do you mean to say that the box is not displaced at all (i.e., is it lying at the same position at the end)?
Obviously no!

But since you've not changed the kinetic energy of the box, nor its potential energy, it makes no difference to the energy where in the plane of reference you've placed it.
 
  • #9


In real sense, however it is not possible to displace a box without doing work even at constant velocity because friction comes into picture.
 

What is the net work done on the box?

The net work done on the box is the total amount of energy that is transferred to or from the box due to outside forces. It can be calculated by multiplying the force applied to the box by the distance the box moves in the direction of the force.

How is net work done on the box different from work?

Net work done on the box takes into account all the forces acting on the box, while work only considers the force that is causing the box to move. Net work also takes into account the direction of the displacement, while work does not.

What factors affect the net work done on the box?

The net work done on the box is affected by the magnitude and direction of the force applied to the box, as well as the distance the box moves in the direction of the force. Other factors such as friction and air resistance may also affect the net work done on the box.

Can the net work done on the box be negative?

Yes, the net work done on the box can be negative if the force applied to the box is in the opposite direction of the displacement. This means that energy is being taken away from the box rather than being transferred to it.

How is the net work done on the box related to the box's kinetic energy?

The net work done on the box is equal to the change in the box's kinetic energy. If the net work done on the box is positive, the box's kinetic energy will increase, and if it is negative, the box's kinetic energy will decrease.

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