Va chạm, công và năng lượng dao động
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| work–energy theorem/wɜːk ˈenədʒi ˈθɪərəm/ | The rule that net work done on a body equals its change in kinetic energy. |
| internal energy/ɪnˈtɜːnl ˈenədʒi/ | The total microscopic kinetic and potential energy of a system's particles, excluding its overall motion and position as a whole. |
| amplitude/ˈæmplɪtjuːd/ | The maximum displacement from an equilibrium position. |
Two directions, one wreck
- A 2 kg cart moving east at 3 m/s and a 1 kg cart moving north at 3 m/s collide and stick. Momentum survives; kinetic energy does not.
- GRE mechanics tests vector momentum, the work–energy theorem 动能定理, spring energy and oscillator energy.
Momentum by components
- Conserve each component separately during the brief collision; take the magnitude only at the end.
$\mathbf p = (6, 0) + (0, 3) = (6, 3)$ kg·m/s, so $\mathbf v = (2, 1)$ m/s. Kinetic energy: $9 + 4.5 = 13.5$ J before, $\tfrac12 \cdot 3 \cdot 5 = 7.5$ J after; 6 J becomes internal energy 内能.
Two 1 kg carts, one at (4, 0) m/s and one at (0, 4) m/s, collide and stick. Final speed in m/s (two decimal places)?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Total momentum is $(4, 4)$ kg m/s shared by 2 kg, so the velocity is $(2, 2)$ m/s and the speed is $2\sqrt2 = 2.83$ m/s.
Kinetic energy lost (J) in that collision?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Before, the kinetic energy is $2 \times \tfrac12 \cdot 1 \cdot 16 = 16$ J. After, it is $\tfrac12 \cdot 2 \cdot 8 = 8$ J, so 8 J is lost.
Work–energy and springs
- $\Delta K = \int \mathbf F_{\text{net}} \cdot d\mathbf r$. Doubling speed quadruples $K$.
- Spring: $U = \tfrac12 kx^2$. Same force $F$: $x = F/k$, $U = \dfrac{F^2}{2k}$ — the softer spring stores more. Same extension: the stiffer one stores more.
Which spring stores more?
Decide which spring stores more energy given what is held fixed.
The same 10 N force stretches spring A ($k = 100$ N/m) and spring B ($k = 400$ N/m). Which stores more energy?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Stored energy is $F^2/2k$ for a fixed force: $100/200 = 0.5$ J for A and $100/800 = 0.125$ J for B. The softer spring stores more.
Oscillator energy
- $E = \tfrac12 mv^2 + \tfrac12 kx^2 = \tfrac12 kA^2 = \tfrac12 mv_{\max}^2$. So $v_{\max} = \omega A$, $\omega = \sqrt{k/m} = 2\pi f$.
- A 0.5 kg oscillator passing equilibrium at 0.2 m/s has $E = 0.01$ J = 10 mJ.
A 0.5 kg mass on a 50 N/m spring oscillates with amplitude 0.1 m. Maximum speed in m/s?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$\omega = \sqrt{k/m} = \sqrt{50/0.5} = 10$ rad/s, and the maximum speed is $\omega A = 10 \times 0.1 = 1$ m/s.
Pendulums and small angles
- $\ddot\theta + \tfrac{g}{L}\sin\theta = 0$; for small $\theta$, $T = 2\pi\sqrt{L/g}$, independent of mass and amplitude 振幅.
- Four times the length doubles the period. Large amplitudes lengthen $T$ slightly.
Doubling the mass of a simple pendulum bob doubles its period.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The period $2\pi\sqrt{l/g}$ contains no mass, because the mass cancels between force and inertia.
Recap
- Add momentum vectors; sticking collisions lose kinetic energy.
- Fixed force: softer spring stores more; fixed extension: stiffer stores more.
- $E = \tfrac12 kA^2$, $v_{\max} = \omega A$; $T = 2\pi\sqrt{L/g}$.
- The work–energy theorem is the rule that net work done on a body equals its change in kinetic energy.