Work done lifting an object 1m (simple question)

In summary, the amount of work done in lifting a ball from one to two metres above the ground is determined by the force exerted by the hand and the distance the ball is lifted. The formula W=Fd applies, where F represents the force exerted by the hand and d represents the distance lifted. It is important to note that this work is only done when there is a change in distance between the starting and ending position, and the displacement value is necessary for unit conversion.
  • #1
femmefatale8
3
0
Hello,

I have a question that should be pretty simple, but I can struggling to understand:

What determines the amount of work done in lifting a ball from one to two metres above the ground?

I know that W=Fd and F=ma , so I would assume that the work done would equal m*a*d.

Something tells me that the answer is not that simple... any help would be appreciated!
 
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  • #2
F=ma only holds for F the NET force. In this case, there are 2 forces, your hand, and gravity. The net force is zero because you should assume that you are lifting the ball at a constant velocity. So F=ma=0 so a=0.

On the other hand, the force in W=Fd can refer to which ever force individually. In this sense, the F in W=Fd is not really the same as the F in F=ma. In this case, you want F to be the force that your hand exerts on the ball, because you want to know the work done by you on the ball.
 
  • #3
Work represents a vector dot product between F(force) and d(distance) so effectively you can say that work is only done when there is a distance changed between the origin of the position and the final position.
For example if I lift a book 1 meter off the ground and exert 10 Newtons to do so then I have done 10 J of work, the change in energy, in this case potential energy.
However, if I were to push the book 1 meter across the ground and exert 10 Newtons to do so then I have not done any work because I have not changed the net energy in the system, thus no change in energy, so no work is done.
 
  • #4
fisixC said:
Work represents a vector dot product between F(force) and d(distance) so effectively you can say that work is only done when there is a distance changed between the origin of the position and the final position.
For example if I lift a book 1 meter off the ground and exert 10 Newtons to do so then I have done 10 J of work, the change in energy, in this case potential energy.
However, if I were to push the book 1 meter across the ground and exert 10 Newtons to do so then I have not done any work because I have not changed the net energy in the system, thus no change in energy, so no work is done.

The second example is not right because in that case the book would acquire kinetic energy. The work you do is still 10J.
 
  • #5
Matterwave said:
The second example is not right because in that case the book would acquire kinetic energy. The work you do is still 10J.

Assuming it took me one second to move the book across the floor.
 
  • #6
Er, it doesn't matter how long it took you to move the book across the floor...W=Fd and as long as d and F are non-zero and not perpendicular to each other then you do some work.
 
  • #7
However you said that "the book would acquire kinetic energy" so by using:
KE = .5m(v^2)
Then without velocity the book wouldn't be able to acquire any kinetic energy.
 
  • #8
Thank you, Matterwave. So, just to clear it up, the answer would simply be:

The amount of work done is determined only by the amount of force my hand exerts? Would displacement be irrelevant on on the final answer, seeing how it is 1?
 
  • #9
By putting a force on the book, you necessarily give it kinetic energy (moving it across a smooth floor). If you want to say that friction has dissipated the kinetic energy, then you do some positive work, and friction does some negative work and the NET work comes out to be 0...but you are still doing work in that case.
 
  • #10
femmefatale8 said:
Thank you, Matterwave. So, just to clear it up, the answer would simply be:

The amount of work done is determined only by the amount of force my hand exerts? Would displacement be irrelevant on on the final answer, seeing how it is 1?

Since you want to find the amount of work done by your hands, then it is only dependent on the force you exert, and the distance. In this question d=1m so when you multiply, you get X Newtons*1m = X joules. The displacement, even though it is 1 meter, is relevant! It gives you the unit conversion correctly!
 

1. What is work?

Work is defined as the product of force and distance, or the force applied to an object multiplied by the distance the object moves in the direction of the force.

2. How is work done when lifting an object 1m?

Work is done when lifting an object 1m by applying a force equal to or greater than the weight of the object, and moving it a distance of 1m in the direction of the force.

3. Is the work done dependent on the weight of the object?

Yes, the work done is directly proportional to the weight of the object. The greater the weight, the more work is required to lift the object 1m.

4. Does the speed at which the object is lifted affect the work done?

No, the work done is independent of the speed at which the object is lifted. As long as the object is lifted a distance of 1m, the work done will be the same regardless of the speed.

5. Can the work done be negative when lifting an object 1m?

Yes, the work done can be negative if the force applied to the object is in the opposite direction of the displacement. This means that energy is being transferred from the object to the surroundings, rather than being added to the object.

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