Error in equation for kinetic energy

In summary, the conversation discusses the relationship between force, distance, and acceleration for an object starting from rest and being pushed by a constant force. The conclusion is that the work done, F*s, is equal to 1/2mv^2 instead of 2mv^2 as previously assumed. This is due to the incorrect assumption that s = v*Δt instead of s = vave*Δt. It is also clarified that for constant acceleration starting from rest, the final velocity is equal to half the average velocity, not twice as previously thought.
  • #1
Hymne
89
1
Okey, let's 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?
 
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  • #2
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
Ah, dt says that t_2 - t_1 approaches zero. But let's 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
Did you know that

a=dv/dt?
 
  • #5
Hymne said:
Okey, let's 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?

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
Hymne said:
Okey, let's 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
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

We also know that when a is constant (which it is due to the constant force) we got:
a*dt = 2v
No, for constant acceleration, starting from rest: a*Δt = v.
which gives us that the work, F*s, equals
2mv^2 instead of (1/2)mv^2
Nope, you get:
F*Δs = ma*Δs = ma*(v/2)*Δt = m*(v/2)*a*Δt = 1/2mv^2.

As expected. :wink:
 

FAQ: Error in equation for kinetic energy

What is the equation for kinetic energy?

The equation for kinetic energy is KE = 1/2 * m * v^2, where KE represents kinetic energy, m represents mass, and v represents velocity.

What are the units for kinetic energy?

The units for kinetic energy are Joules (J).

What causes errors in the equation for kinetic energy?

Errors in the equation for kinetic energy can be caused by incorrect values for mass or velocity, using the wrong units, or not accounting for external forces such as friction.

Can kinetic energy be negative?

No, kinetic energy cannot be negative. It is always a positive value representing the energy of motion.

How does kinetic energy relate to potential energy?

Kinetic energy and potential energy are both forms of energy and can be interconverted. Kinetic energy is the energy of motion, while potential energy is the energy that an object has due to its position or state. In some cases, potential energy can be converted into kinetic energy and vice versa, such as in the case of a ball rolling down a hill.

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