# Question on energy

1. Jul 3, 2007

### ray4758026

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 isnt work defined as Ft??

2. Jul 3, 2007

3. Jul 3, 2007

### cristo

Staff Emeritus
It's a definition. That's like asking "why is acceleration defined as the rate of change of velocity?"
Why do you think that?
Because that is a different quantity.

4. Jul 3, 2007

### Staff: Mentor

No. The change in kinetic energy will be the same for each.
Ft already has a name: Impulse. And it equals the change in momentum, not energy.

5. Jul 3, 2007

### K.J.Healey

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).

6. Jul 3, 2007

### mgb_phys

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?