Work done lifting an object 1m (simple question)


by femmefatale8
Tags: displacement, lift, work
femmefatale8
femmefatale8 is offline
#1
Mar1-11, 07:58 PM
P: 3
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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Matterwave
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#2
Mar1-11, 08:04 PM
P: 2,043
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.
fisixC
fisixC is offline
#3
Mar1-11, 08:16 PM
P: 54
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.

Matterwave
Matterwave is offline
#4
Mar1-11, 08:19 PM
P: 2,043

Work done lifting an object 1m (simple question)


Quote Quote by fisixC View Post
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.
fisixC
fisixC is offline
#5
Mar1-11, 08:21 PM
P: 54
Quote Quote by Matterwave View Post
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.
Matterwave
Matterwave is offline
#6
Mar1-11, 08:24 PM
P: 2,043
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.
fisixC
fisixC is offline
#7
Mar1-11, 08:27 PM
P: 54
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.
femmefatale8
femmefatale8 is offline
#8
Mar1-11, 08:29 PM
P: 3
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?
Matterwave
Matterwave is offline
#9
Mar1-11, 08:30 PM
P: 2,043
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.
Matterwave
Matterwave is offline
#10
Mar1-11, 08:32 PM
P: 2,043
Quote Quote by femmefatale8 View Post
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!


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