Why is Work Defined as Force X Distance? Explained

  • Thread starter Thread starter ray4758026
  • Start date Start date
  • Tags Tags
    Energy
Click For Summary
Work is defined as force multiplied by distance (Fd) because it quantifies the energy transferred when a force is applied over a distance. Applying the same force to different masses over the same distance results in the same work done, as the change in kinetic energy is equal for both masses. The term Ft refers to impulse, which measures the change in momentum, not energy. In scenarios with friction, more force may be needed for heavier masses, but the fundamental definition of work remains unchanged. Ultimately, work as Fd captures the relationship between force and distance without considering the speed of application.
ray4758026
Messages
4
Reaction score
0
Why is work defined as Fd? If you apply the same force to 2 different masses over the same distance surely the larger mass has more energy. Why isn't work defined as Ft??
 
Physics news on Phys.org
what is your 'F'?
 
ray4758026 said:
Why is work defined as Fd?
It's a definition. That's like asking "why is acceleration defined as the rate of change of velocity?"
If you apply the same force to 2 different masses over the same distance surely the larger mass has more energy.
Why do you think that?
Why isn't work defined as Ft??
Because that is a different quantity.
 
ray4758026 said:
If you apply the same force to 2 different masses over the same distance surely the larger mass has more energy.
No. The change in kinetic energy will be the same for each.
Why isn't work defined as Ft?
Ft already has a name: Impulse. And it equals the change in momentum, not energy.
 
Just so you can see the Kinetic Energy would be the same for each (as well as work done, obviously)
F=F1=F2=m1*a1=m2*a2 (standard definition for some blocks on a flat surface)
a1=F/m1 (accel)
dv/dt = F/m1 (accel in differential)
dv/dx dx/dt = F/m1 (product rule)
v*(dv/dx) = F/m1 (velocity def)
Integral(v*dv) = Integral (F/m1 * dx) (v=0..v,x=0..x)
1/2 v^2 = (F/m1) * x
(1/2)*m1*v^2 = F*x == Kinetic Energy = Work Done given some distance X from a point at which V=0.

Same with F2=m2a2, youll get F*x
so the kinetic energies are the same. F*x (Fd).
 
ray4758026 said:
Why is work defined as Fd? If you apply the same force to 2 different masses over the same distance surely the larger mass has more energy. Why isn't work defined as Ft??

This is one of those real life experience meets physics questions.
If you are pushing the two masses along a table with no friction with the same force then the larger mass will accelerate less but the work done will be the same. Heavy mass moves slowly = small mass moved quickly = same energy!

If you have a 'real' table then it will take more work to move the heavier mass but this is because of the greater friction so you put in more force and hence more energy. Or alternatively the heavier mass moved even more slowly because you 'used' some more of the force to beat friction.
In either case same force, same distance = same energy.

If you are lifting the masses the same distance vertically then you have to put more force to the heavier mass and so use more energy. A heavier weight stored at the same height obviously has more (potential) energy then a light one.

The reason it is Fd not Ft is that imagine you quickly lifted a weight like a weightlifter and then lifted another identical weight slowly by pushing it up a ramp. If you came back later would you expect the quicker lifted weight to have more potential energy because it remembered how quickly it had got there?
 
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 18 ·
Replies
18
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 12 ·
Replies
12
Views
4K
Replies
15
Views
2K