Register to reply 
Work done lifting an object 1m (simple question) 
Share this thread: 
#1
Mar111, 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! 


#2
Mar111, 08:04 PM

Sci Advisor
P: 2,811

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
Mar111, 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. 


#4
Mar111, 08:19 PM

Sci Advisor
P: 2,811

Work done lifting an object 1m (simple question)



#5
Mar111, 08:21 PM

P: 54




#6
Mar111, 08:24 PM

Sci Advisor
P: 2,811

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 nonzero and not perpendicular to each other then you do some work.



#7
Mar111, 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. 


#8
Mar111, 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? 


#9
Mar111, 08:30 PM

Sci Advisor
P: 2,811

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
Mar111, 08:32 PM

Sci Advisor
P: 2,811




Register to reply 
Related Discussions  
Does lifting an object higher increase its potential energy?  Classical Physics  12  
Work lifting cable and work pumping water  Introductory Physics Homework  1  
Work done lifting an object underwater  Classical Physics  7  
Is work done in lifting an object with a spring more than lifting it with a rod?  Classical Physics  6  
A hypothetical pressure question.. lifting a 10,000kg object  Introductory Physics Homework  1 