# Error in equation for kinetic energy

1. Jan 7, 2009

### Hymne

Okey, lets say we got an object with no start velocity (V_i = 0) which is being pushed by a constant force F, under a certain distance s (during a time intervall dt).
We get through F = ma that
F*s = ma*s = ma*v*dt
We also know that when a is constant (which it is due to the constant force) we got:
a*dt = 2v
which gives us that the work, F*s, equals
2mv^2 instead of (1/2)mv^2
What is wrong?

2. Jan 7, 2009

### Dr.D

The object does not travel a finite distance s during the infinitesimal time dt. You are mixing your concepts here.

F*ds = m*a*v*dt

Try taking it from here.

3. Jan 7, 2009

### Hymne

Ah, dt says that t_2 - t_1 approaches zero. But lets just use (t_2 - t_1), I still get it wrong :(
F*s = m*a*v*(t_2 - t_1) = m*v*2v = 2mv^2

I assume that over our given time and with our constant force we have a * (t_2 - t_1) = 2*v.

4. Jan 7, 2009

### Dr.D

Did you know that

a=dv/dt?

5. Jan 7, 2009

### Redbelly98

Staff Emeritus
You're using "v" to mean the average velocity (= s/dt)

When we say that work = (1/2) m v^2, "v" is the final velocity.

The final velocity is twice the average velocity, they are not equal.

6. Jan 7, 2009

### Staff: Mentor

You here assume that s = v*Δt, but that should be s = vave*Δt, where vave is the average speed. For constant acceleration starting from rest, vave = v/2, where v is the final speed. So really:
F*Δs = ma*Δs = ma*(v/2)*Δt

No, for constant acceleration, starting from rest: a*Δt = v.
Nope, you get:
F*Δs = ma*Δs = ma*(v/2)*Δt = m*(v/2)*a*Δt = 1/2mv^2.

As expected.