# Does a moving massive object have more kinetic energy than a lighter object?

Jan 21, 2017

Maybe.

#### Explanation:

The problem doesn't specify this, but if are to assume that both the massive object and the lighter object are moving with the same velocity, then the answer would be yes.

An object's kinetic energy is simply the energetic cost paid, i.e. work done on the object, in order to accelerate it from rest to a given velocity, let's say $v$.

Now, an object's mass comes into play here because it will affect the amount of work we do in order to accelerate the object, i.e. give it kinetic energy.

More specifically, a more massive object will require more work in order to get from rest to a velocity $v$, and consequently have a higher kinetic energy, than a lighter object that goes from rest to the same velocity $v$.

Once the object reaches velocity $v$, its kinetic energy can be expressed as a function of its velocity and of its mass

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{kinetic energy" = 1/2 xx "mass" xx "velocity}}^{2}}}}$

If you take $M$ to be the mass of the massive object and $m$ to be the mass of the lighter object, you can say that

${\text{KE"_"massive" = 1/2 * M * v_"massive}}^{2}$

${\text{KE"_ "light" = 1/2 * m * v_"light}}^{2}$

If you take

${v}_{\text{massive" = v_"light}} \to$ the two objects are moving with the same velocity

then you can say that

"KE"_ "massive"/"KE"_ "light" = (color(red)(cancel(color(black)(1/2))) * M * color(red)(cancel(color(black)(v^2))))/(color(red)(cancel(color(black)(1/2))) * m * color(red)(cancel(color(black)(v^2)))

which is equivalent to

$\text{KE"_"massive" = M/m * "KE"_"light}$

If $M > m$, you have $\frac{M}{m} > 1$ and

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{KE"_ "massive" > "KE"_"light}}}}$

This means that when two objects are moving with the same velocity, the object that has the greater mass will have the higher kinetic energy

Let's take a numerical example to illustrate this idea. Let's assume that you have a bicycle of mass $m = \text{10 kg}$ and a car of mass $M = \text{1000 kg}$, and that both of them are moving at ${\text{10 m s}}^{- 1}$.

You will have

$\text{KE"_"car" = 1/2 * "1000 kg" * "10 m s"^(-1) = "5000 J}$

$\text{KE"_"bicycle" = 1/2 * "10 kg" * "10 m s"^(-1) = "50 J}$

Notice that

"KE"_ "car"/"KE"_ "bicyle" = (5000 color(red)(cancel(color(black)("J"))))/(50color(red)(cancel(color(black)("J")))) = 100

The car has a kinetic energy is $100$ times higher than the kinetic energy of the bicycle, which means that it took $100$ times more work to accelerate the car from rest to ${\text{10 m s}}^{- 1}$ than it took to accelerate the bicycle from rest to ${\text{10 m s}}^{- 1}$.

You can test this yourself by trying to push a bicycle and a car to the same velocity. The one that is harder to push will have the higher kinetic energy once it gets to that velocity.

$\textcolor{red}{\underline{\textcolor{b l a c k}{\text{more work needed to accelerate to the same velocity " = " higher kinetic energy}}}}$