Derivations of KE=1/2mv^2 and PE=mgh

In summary, the derivations for gravitational potential energy and kinetic energy without using calculus involve understanding energy as the ability to do work. For potential energy, this is measured by the force (mg) multiplied by the distance through which the force acts (h). For kinetic energy, it is measured by the work done by a body traveling at velocity v before it stops, which is equal to the force (mv) multiplied by the average distance traveled (vavgΔt).

Homework Statement

I need the derivations of gravitational potential energy and kinetic energy (without using calculus)

Homework Equations

$PE = mgh \\ KE = {1 \over 2}mv^2$

The Attempt at a Solution

Potential Energy:

Work done = $Fs$
But... $F=ma$
So substituting that in becomes $mas$
where a is the acceleration due to gravity (the g in mgh), and s is the height in mgh.Kinetic Energy (this is I am a lot less sure on):

Work done = $Fs$
But... $F=ma$
So substituting that in becomes $mas$
And when u=0 $v^2 = 0 + 2as$
Which can be arranged to $a = {v^2 \over 2s}$
So substituting that in becomes $= m \times {v^2 \over 2s} \times s$
Which can apparently (so I was told) be rearranged to become ${1 \over 2}mv^2$

Can anyone let me know if these are correct, if not any advice is appreciated and if they are correct, any advice on how to word it correctly is also appreciated.

Thanks.

Homework Statement

I need the derivations of gravitational potential energy and kinetic energy (without using calculus)

Homework Equations

$PE = mgh \\ KE = {1 \over 2}mv^2$

The Attempt at a Solution

Potential Energy:

Work done = $Fs$
But... $F=ma$
So substituting that in becomes $mas$
where a is the acceleration due to gravity (the g in mgh), and s is the height in mgh.
It is fine but your explanation could be clearer. Start with the definition of Energy as "the ability to do work". The amount of work a body of mass m will do in falling from point A to B through a vertical distance h is a measure of its potential energy at A relative to B. That amount of work is the force ( = mg) multiplied by the distance through which the force acts (h).

Kinetic Energy (this is I am a lot less sure on):t
This you need a bit of help on. Start with the same definition of energy. The kinetic energy is the ability to do work due to a body's motion rather than its position.

How much work will a body traveling at velocity v do before it stops? That is a measure of its kinetic energy.

Use:

W = ΔKE = FΔs
FΔt = -mΔv = m(v-0) = mv
Δs = vavgΔt

AM

Last edited:

1. What is the meaning of KE=1/2mv^2 and PE=mgh?

KE=1/2mv^2 and PE=mgh are two equations used in physics to calculate the kinetic energy and potential energy of an object. KE stands for kinetic energy, which is the energy an object possesses due to its motion. PE stands for potential energy, which is the energy an object possesses due to its position or state. These equations help us understand the energy of an object and how it changes in different situations.

2. How do you derive the equation KE=1/2mv^2?

The equation KE=1/2mv^2 can be derived using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. By integrating the equation for work, W=Fd, and using the definition of acceleration, a=dv/dt, we can derive the equation KE=1/2mv^2.

3. How do you derive the equation PE=mgh?

The equation PE=mgh can be derived using the principles of work and energy. The work done on an object by a force is equal to the change in its potential energy. By integrating the equation for work, W=Fd, and using the definition of gravitational force, F=mg, we can derive the equation PE=mgh.

4. What does each variable in the equations represent?

In the equation KE=1/2mv^2, m represents the mass of the object and v represents its velocity. In the equation PE=mgh, m represents the mass of the object, g represents the acceleration due to gravity, and h represents the height of the object.

5. Can these equations be applied to all objects?

Yes, these equations can be applied to all objects as long as they have mass and are experiencing motion and/or have a position or state relative to a gravitational field. However, these equations are most accurate for objects that are moving at non-relativistic speeds and are not subject to extreme gravitational forces.

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