| English | 意味 |
|---|---|
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | 運動エネルギー |
| scalar/ˈskeɪlə/ | スカラー量 |
| work-energy theorem/wɜːk ˈenədʒi ˈθɪərəm/ | 仕事-エネルギーの定理 |
| net work/net wɜːk/ | 正味の仕事 |
Why speed is so dangerous
- A car at $40\ \text{km/h}$ does far more than double the damage of one at $20\ \text{km/h}$.
- It does four times the damage -- because energy depends on the square of speed.
- That hidden square is why small speed increases matter so much.
- The energy of motion has a name and a formula.
Kinetic energy
- Kinetic energy 动能 is the energy an object has because it is moving:
- It is a scalar 标量 -- just a number, with no direction.
- Bigger mass or bigger speed means more kinetic energy.
A $4\ \text{kg}$ object moves at $3\ \tfrac{\text{m}}{\text{s}}$. What is its kinetic energy (in J)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$K = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(4)(9) = 18\ \text{J}$.
Kinetic energy can be negative if an object moves backward.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The $v^2$ makes $K$ positive regardless of direction -- kinetic energy is a scalar and never negative.
The square is everything
- Because of the $v^2$, doubling the speed quadruples the kinetic energy.
- Triple the speed and you get nine times the energy.
- Mass matters too, but only linearly -- speed is the powerful one.
If a car's speed doubles, its kinetic energy becomes...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$K \propto v^2$, so doubling $v$ multiplies $K$ by $2^2 = 4$.
Select all true statements about kinetic energy.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Kinetic energy is a positive scalar that scales with $v^2$; it is never negative.
The work-energy theorem
- The work-energy theorem 功能定理 links force to energy:
- The net work done on an object equals its change in kinetic energy.
- Positive net work speeds it up; negative net work slows it down.
Kinetic energy grows with speed squared
Double the speed and the kinetic energy quadruples - it depends on the square of speed.
A $3\ \text{kg}$ cart goes from $2$ to $4\ \tfrac{\text{m}}{\text{s}}$. How much net work was done on it (in J)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$\Delta K = \tfrac{1}{2}(3)(4^2) - \tfrac{1}{2}(3)(2^2) = 24 - 6 = 18\ \text{J}$.
The work-energy theorem says the net work equals the change in ____ energy.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$W_{net} = \Delta K$ -- net work in, kinetic energy up.
For a whole system
- For a group of particles, add up each one's $\tfrac{1}{2}mv^2$ to get the total kinetic energy.
- The work-energy theorem still holds for the system as a whole.
- This lets energy handle complicated motions a single force diagram cannot.
A $2\ \text{kg}$ ball speeds up from $3$ to $5\ \tfrac{\text{m}}{\text{s}}$.
- $K_i = \tfrac{1}{2}(2)(3^2) = 9\ \text{J}$; $K_f = \tfrac{1}{2}(2)(5^2) = 25\ \text{J}$.
- By the theorem, the net work done was $\Delta K = 25 - 9 = 16\ \text{J}$.
Kinetic energy is never negative -- the $v^2$ makes it positive whichever way the object moves. And doubling speed does not double the energy; it quadruples it. Reaching for "twice as fast, twice the energy" is the classic trap.
Kinetic energy $K = \tfrac{1}{2}mv^2$ is a scalar that grows with the square of speed -- double the speed, quadruple the energy. The work-energy theorem says the net work equals the change in kinetic energy, $W_{net} = \Delta K$.