Translational Kinetic Energy
| English | Chinese | Pinyin |
|---|---|---|
| kinetic energy | 动能 | dòng néng |
| joule | 焦耳 | jiāo ěr |
Twice the speed, four times the danger
- A car at $60\ \tfrac{\text{km}}{\text{h}}$ and the same car at $120\ \tfrac{\text{km}}{\text{h}}$ — twice the speed.
- But the fast car carries four times the energy of motion, not twice.
- That energy is kinetic energy 动能, and it grows with the square of speed.
- It is why crashes at high speed are so much more destructive.
The kinetic energy formula
- A moving mass stores kinetic energy $E_k = \tfrac12 m v^2$.
- It is a scalar (no direction) measured in joules 焦耳 ($\text{J}$).
- It depends on mass $m$ and speed $v$ — and always positive for a moving object.
- At rest ($v = 0$) the kinetic energy is zero.
What is the kinetic energy of a $2\ \text{kg}$ ball moving at $3\ \tfrac{\text{m}}{\text{s}}$, in joules?
$E_k = \tfrac12 m v^2 = \tfrac12 (2)(9) = 9\ \text{J}$.
Kinetic energy is a vector, so it has a direction.
Kinetic energy is a scalar — a number of joules with no direction.
Why $v^2$ matters so much
- Because of the $v^2$, doubling the speed makes $E_k$ four times larger.
- Tripling the speed makes it nine times larger.
- So a small speed increase is a large energy (and braking-distance) increase.
- Mass matters too, but only in direct proportion — double $m$, double $E_k$.
Energy grows as v squared
E_k = a·v² (here a = ½m)
Trace the quadratic curve of kinetic energy against speed and see how fast it climbs.
If a car doubles its speed, its kinetic energy becomes:
$E_k \propto v^2$, so doubling $v$ gives $2^2 = 4$ times the kinetic energy.
The same $2\ \text{kg}$ ball now moves at $6\ \tfrac{\text{m}}{\text{s}}$. What is its kinetic energy, in joules?
$E_k = \tfrac12 (2)(6^2) = 36\ \text{J}$ — four times the $9\ \text{J}$ at $3\ \tfrac{\text{m}}{\text{s}}$.
Select all true statements about kinetic energy.
$E_k = \tfrac12 mv^2 \ge 0$: it depends on $v^2$, is in joules, and is zero at rest. It is never negative.
Work changes kinetic energy
- To speed something up you must do work on it (next lesson).
- The work–energy theorem: the net work equals the change in kinetic energy, $W_{\text{net}} = \Delta E_k$.
- Positive net work speeds it up; negative net work (like braking) slows it down.
- Kinetic energy is the "bank account" that work pays into or draws out of.
The work–energy theorem says the net work equals the change in ____ energy.
$W_{\text{net}} = \Delta E_k$ — net work changes the kinetic energy.
Kinetic energy depends on $v^2$, not $v$. A common error is to think doubling the speed doubles the energy. It quadruples it — which is why stopping distances grow so quickly with speed.
Find the kinetic energy of a $2\ \text{kg}$ ball moving at $3\ \tfrac{\text{m}}{\text{s}}$.
- $E_k = \tfrac12 m v^2 = \tfrac12 (2)(3^2) = \tfrac12 (2)(9) = 9\ \text{J}$.
Double the speed to $6\ \tfrac{\text{m}}{\text{s}}$ and $E_k = \tfrac12(2)(36) = 36\ \text{J}$ — four times as much.
Kinetic energy is the energy of motion, $E_k = \tfrac12 m v^2$ (a scalar, in joules). It depends on the square of speed, so doubling $v$ quadruples $E_k$. The work–energy theorem links it to work: $W_{\text{net}} = \Delta E_k$.