# Confused by the relationship of work and kinetic energy

• Oleiv
It's not magic. It's just the result of doing an integral.In summary, the conversation discusses the equation for work and its relation to units of measurement, specifically in regards to the quantity of kinetic energy. It is noted that the 1/2 factor in the equation for kinetic energy is a conversion factor and does not affect the dimensions or units in the equation. It is also mentioned that in the equation for work, the speed will change while work is applied, whereas in the equation for kinetic energy, the final speed is used. The conversation also emphasizes the difference between dimension analysis and integration in terms of solving equations.

#### Oleiv

So the equation for work is W = F * s
F = m * a, so W = m * a * s
Transferring this to units of measurement gives us: J = kg * m * s-2 * m
Or simplified: J = kg * m2 * s-2
Transferring back to units of quantity: W = m * v2
How can that be correct? Obviously Ekin = 1/2 * m * v2. Where did that 1/2 go? Or is W =/= Ekin? Am I making some other kind of mistake?

Oleiv said:
Where did that 1/2 go?
It's the rules of Integration. Integrate xdx and you get x2/2
Graphically, it's the area of the v/t triangle or the F/x triangle etc etc.

• DrClaude
Oleiv said:
So the equation for work is W = F * s
F = m * a, so W = m * a * s
Transferring this to units of measurement gives us: J = kg * m * s-2 * m
Or simplified: J = kg * m2 * s-2
Transferring back to units of quantity: W = m * v2
How can that be correct? Obviously Ekin = 1/2 * m * v2. Where did that 1/2 go? Or is W =/= Ekin? Am I making some other kind of mistake?

The ##\frac12## is effectively a conversion factor based on your choice of units. In any case, ##\frac12## is dimensionless and doesn't affect the dimensions or units in an equation.

For example, the area of a circle is ##A = \pi r^2##. Both quantities have dimensions of ##L^2## or SI units of ##m^2##. ##\pi## is a dimensionless, unit-less factor based on the geometry of the circle.

• DrClaude
Oleiv said:
Transferring back to units of quantity: W = m * v2
How can that be correct? Obviously Ekin = 1/2 * m * v2. Where did that 1/2 go? Or is W =/= Ekin? Am I making some other kind of mistake?
To add a somewhat more physical explanation to the very good posts of @sophiecentaur and @PeroK, when you write the ##v## in the two equations is not the same. In ##E_\mathrm{kin}##, it is the final speed, when the work has stopped, while in the equation for work, the speed will change while work is applied. In the case of constant acceleration, we can write ##W = m \langle v \rangle^2 = mv_f^2/2##, with ##\langle v \rangle## the average speed while the work is applied.

Edit: this assumes also that the speed was initially 0. Otherwise, we would have to write the equations in terms of ##\Delta v## and ##\Delta E_\mathrm{kin}##.

Oleiv said:
So the equation for work is W = F * s
F = m * a, so W = m * a * s
Transferring this to units of measurement gives us: J = kg * m * s-2 * m
Or simplified: J = kg * m2 * s-2
Transferring back to units of quantity: W = m * v2
How can that be correct? Obviously Ekin = 1/2 * m * v2. Where did that 1/2 go? Or is W =/= Ekin? Am I making some other kind of mistake?
Dimension analysis is not the same as doing integration (or algebra if you assume constant acceleration). That's the mistake.

At constant acceleration, s=1/2at2

That's where the 1/2 comes from.

## 1. What is work and kinetic energy?

Work is the measure of the amount of force applied to an object over a certain distance. Kinetic energy is the energy an object possesses due to its motion.

## 2. How are work and kinetic energy related?

Work and kinetic energy are directly related. Work is responsible for changing an object's kinetic energy. When work is done on an object, its kinetic energy increases. Similarly, when work is done by an object, its kinetic energy decreases.

## 3. What is the difference between work and kinetic energy?

The main difference between work and kinetic energy is that work is a measure of the amount of force applied to an object, while kinetic energy is a measure of the object's motion. Work can be positive or negative, while kinetic energy is always positive.

## 4. How is the relationship between work and kinetic energy represented mathematically?

The relationship between work and kinetic energy is represented by the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy. This can be mathematically expressed as W = ΔKE = KEf - KEi, where W is work, ΔKE is the change in kinetic energy, KEf is the final kinetic energy, and KEi is the initial kinetic energy.

## 5. Why is understanding the relationship between work and kinetic energy important?

Understanding the relationship between work and kinetic energy is important because it helps us understand the factors that affect an object's motion and how to manipulate them. It also allows us to calculate the amount of work and energy involved in various situations, which is crucial in many scientific and engineering applications.