Conservation of Energy when lifting a box up off the floor

In summary, your problem is that you are not accounting for the work done by the force on the system.
  • #1
nav888
38
2
Homework Statement
Writing a conservation of energy statement
Relevant Equations
W = mgh
E = 1/2 mv^2
So, I cannot for the life of me write a conservation of energy statement, when an object is lifted up by a force. So in my example there is a box on the floor with v = 0, and then a force of magnitude F, where F > mg, acts on the ball, now the net force is F-mg, and hence the work done is (F - mg) h where h is the vertical height, Now I thought that the work done is equal to the energy of the ball, and hence (F - mg) h = 1/2mv^2 + mgh.
But this isnt correct, could someone please help explain?

Many thanks
 
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  • #2
Please read your post and correct ...
 
  • #3
BvU said:
Please read your post and correct ...
? I have come here for help I cannot correct my work or I would not ask for help? Sorry
 
  • #4
@nav888 I hope I understood your problem correctly.
##W_{total}=\Delta K=\frac 1 2 m(v^2_f-v^2_i)##
In your example ##v_i=0## so the total work is : ##W_{total}=\frac 1 2 mv^2_f##
nav888 said:
(F - mg) h = 1/2mv^2 + mgh.
If there is no air resistance and only these 2 forces are acting on the object you can calculate total work easily:
##W_{total}=F_{net}dcos\theta=(F-mg)h=\frac 1 2 mv^2_f##

Your answer has an extra mgh:
(F - mg) h = 1/2mv^2 + mgh.

Edit: Note that The work-energy theorem states that the total work done on an object equals the change in its kinetic energy.
 
  • #5
What you need is the work-energy theorem because mechanical energy (kinetic plus potential) is not conserved. The work-energy theorem says that the sum of the works done by all the forces is equal to the change in kinetic energy. If (constant) external force ##F## lifts the mass to height ##h## from rest, there are two forces doing work on the mass, gravity and ##F##. So you write
##W_g=-mgh## (negative work because the displacement and the force are in opposite directions)
##W_F=Fh## (positive work because the displacement and the force are in the same direction)
##W_{total}=W_g+W_F=Fh-mgh##
Thus, by the work-energy theorem $$\Delta K = \frac{1}{2}mv^2=Fh-mgh.$$Now the change in potential energy is ##\Delta U = mgh.## If you move that to the other side of the equation and change sign, you get $$\Delta K+\Delta U=Fh.$$ The above equation says that the sum of the changes in kinetic plus potential energy changes, as the mass is lifted, is not constant. If mechanical energy were conserved, then the sum of the two changes must be zero.

So you cannot, for the life of you, write an energy conservation equation simply because you chose a situation where energy is not conserved.
 
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  • #6
Another way of looking at it:
Posts #4 and #5 correctly analyse it in terms of the work done on the mass.
Alternatively, consider the work done by the force on the mass/Earth system. The gained potential energy is internal energy of that system:
##Fh=\frac 12mv^2+mgh##.
So your error can be thought of as mixing up the two views.
 
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