Work and energy: the ledger must tie
導入済み| English | 意味 |
|---|---|
| spring constant/sprɪŋ ˈkɒnstənt/ | The stiffness of a linear spring: restoring-force magnitude divided by displacement from its unstretched position. |
| potential energy/pəˈtenʃl ˈenədʒi/ | Energy associated with a system's configuration, such as height in a uniform gravitational field or extension of an ideal spring. |
| work/wɜːk/ | Energy transferred by a force through displacement, calculated by integrating the force component along the displacement. |
| energy method | Using work and energy relations to determine motion without solving the full time-dependent force equations. |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | half m v squared, a scalar with no components and no direction |
Gravity pays, the spring banks
- The mount is a spring of spring constant 弹簧刚度 $k = 4900$ N/m; the drop test is bookkeeping: gravity pays $mg(h + x)$ from its potential energy 势能, the spring banks $\frac{1}{2}kx^2$ - and the ledger must tie.
- The $+x$ is the deflection gravity keeps working through: the term every rushed report drops.
- The energy method 能量法 is exactly this bookkeeping: one equation, blind to time.
*Both ledger sides read 47.56 J - that is the second-way check
The ramp ledger: potential energy released over 1.50 m at 25 degrees for the 2.5 kg till (J, three decimals)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
m g s sin 25°.
In mg(h + x) = k x²/2, the +x exists because:
難易度 1/5 · 100 コイン + 今日初回達成時の小ボーナス
The weight descends h plus the whole compression.
The arrest, without integrating
- $x_{max} = 0.1393$ m, so $F_{max} = kx = 682.7$ N - the peak is twice the average spring force, and no force integral was ever evaluated.
- Deceleration $273.1$ m/s² $= 27.84$ g against a 30 g rating: margin 7.2%.
Where does the joule go?
Route each energy term of the drop test to its store.
A 2.5 kg till dropped 1.80 m onto k = 4900 N/m: maximum deflection (m, four decimals)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
mg(h+x) = k x²/2.
Same drop: peak force (N, one decimal)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
F = k x.
Same drop: peak deceleration in g (two decimals)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
kx / m / 9.81.
The ramp, again, by energy
- $mgs\sin\beta - \mu_k mg\cos\beta \cdot s = \frac{1}{2}mv^2$: $15.547 - 6.668 = 8.879$ J of kinetic energy 动能 gives $v = 2.665$ m/s.
- The same speed Newton's integration gave - two routes, one number.
Drop from only 0.900 m instead: peak force (N, one decimal)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
x = 0.1001 m this time - the peak does not halve.
With friction taking 6.668 J: kinetic energy at the end (J, three decimals)?
難易度 2/5 · 140 コイン + 今日初回達成時の小ボーナス
15.547 - 6.668.
Kinetic energy needs components along each axis to be added correctly.
難易度 1/5 · 100 コイン + 今日初回達成時の小ボーナス
It is a scalar: half m v squared with the total speed.
Three energy slips
- Dropping the +x. Gravity works through the whole deflection.
- Counting gravity twice. A force on the left or a potential on the right - never both.
- Asking energy for a time. Energy answers how fast and how far; time is integration's job.
Recap
- $T_1 + V_1 + U_{1\to2} = T_2 + V_2$; springs at $\frac{1}{2}kx^2$.
- Drop test: $x = 0.1393$ m, 682.7 N, 27.84 g, ledger 47.56 = 47.56 J.
- Ramp: $15.547 - 6.668 = 8.879$ J → $2.665$ m/s.