Prove Kinetic Energy equation (without calculus)

[Chrystalle]

Homework Statement

Prove kinetic energy is relationship between mass and velocity using E(g)=mgh (no calculus, momentum, kinematics).

Homework Equations

gravitational energy: E=mgh
kinetic energy: E=(1/2)mv^2

The Attempt at a Solution

I know it can be derived using the gravitational energy equation (E=mgh) and the kinematic equation (Δv^2)=2aΔx.

If I'm correct, conservation of energy (in free fall situation) states that E gravitational before being dropped = E kinetic when object hits the ground. I've been led to believe that I can prove the kinetic energy formula only needing E(g)=mgh and E(g)=E(k). This does not seem like enough information though. I'm stuck trying to prove that height doesn't affect kinetic energy and that the relationship between mass and velocity is (1/2) and ^2.

Am I wrong that kinetic energy can be derived only knowing E(g)=mgh and E(g)=E(k)?

 

[Shinaolord]

E , when considering the total energy, isn't a function of k, it's a function of height (The potential part, The sum is obviously a constant), which in turn can be considered a function of time. ##E(h(t))##EDIT: what you are given is the only necessary info.

 

[Chrystalle]

E , when considering the total energy, isn't a function of k, it's a function of height (The potential part, The sum is obviously a constant), which in turn can be considered a function of time. ##E(h(t))##EDIT: what you are given is the only necessary info.

I thought energy was a poor tool for time?

 

[epenguin]

Homework Statement

Prove kinetic energy is relationship between mass and velocity using E(g)=mgh (no calculus, momentum, kinematics).

Homework Equations

gravitational energy: E=mgh
kinetic energy
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Homework Statement

Prove kinetic energy is relationship between mass and velocity using E(g)=mgh (no calculus, momentum, kinematics).

Homework Equations

gravitational energy: E=mgh
kinetic energy: E=(1/2)mv^2

The Attempt at a Solution

I know it can be derived using the gravitational energy equation (E=mgh) and the kinematic equation (Δv^2)=2aΔx.

If I'm correct, conservation of energy (in free fall situation) states that E gravitational before being dropped = E kinetic when object hits the ground. I've been led to believe that I can prove the kinetic energy formula only needing E(g)=mgh and E(g)=E(k). This does not seem like enough information though. I'm stuck trying to prove that height doesn't affect kinetic energy and that the relationship between mass and velocity is (1/2) and ^2.

Am I wrong that kinetic energy can be derived only knowing E(g)=mgh and E(g)=E(k)?

Is this a homework problem? Because it is not very clearly set out what you are allowed to assume, and whether that is in your 2 or whether 2 is your own. If it is a homework problem please quote verbatim.

I think you will probably find it helpful to look back to earlier lessons or textbook sections about kinematics of constant acceleration. Here you are re-doing it with concepts of force, work, potential, and mass added in.

So on the one hand you've got force, mg, X distance moved which is h.
On the other hand this constant force causes a constant acceleration g.
Which is a line of v against t of constant slope.
Now where your particle gets to is time-averaged velocity X time, t.
If the acceleration is constant then the average velocity is half the final velocity, that's where the /2 factor comes from, v_average = v/2, and why you can do it without calculus.
You can manage to eliminate time from this.

- h = v_averaget

v = gt. (constant acceleration)

Combine and get

- h = v_averagev/g

Further combine and get

- gh = v²/2

In the result both sides are multiplied by m. You might think it needn't be there. You'd be right as far as pure gravitational problems are concerned. Gravitational mass = inertial mass.
This depended on constant acceleration, enabling to do without calculus. You could think that as this true for every little 'infinitesimal' part it will be true whe you add them all so conservation energy generally true when force depends only on position, a quasicalculus reasoning.
I am talking about when initial velocity = 0 to be less complicated.

Waves hands. Hope helps.