The first law expresses the principle of conservation of energy for thermodynamic systems. It states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system on its surroundings.
$$\Delta U = Q - W$$For an infinitesimal process the first law can be written as
$$dU = \delta Q - \delta W$$In many A‑Level problems the work term is expressed as $p\,dV$, giving
$$dU = \delta Q - p\,dV$$Oscillatory motion (e.g., a mass‑spring system) involves continual inter‑conversion between kinetic and potential energy. When the oscillator is coupled to a thermal reservoir, the first law governs the energy exchange.
| Quantity | Expression |
|---|---|
| Kinetic Energy ($K$) | $K = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t)$ |
| Elastic Potential Energy ($U_s$) | $U_s = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \cos^2(\omega t)$ |
| Total Mechanical Energy ($E_{\text{mech}}$) | $E_{\text{mech}} = K + U_s = \frac{1}{2} k A^2$ (constant if no dissipation) |
When damping is present (e.g., due to air resistance), mechanical energy is not conserved. The lost mechanical energy appears as heat transferred to the surroundings, and the first law becomes
$$\Delta U_{\text{int}} = Q_{\text{diss}} = \int F_{\text{damp}}\,dx$$In some systems a periodic heat input can sustain oscillations (e.g., a Stirling engine). The first law links the supplied heat $Q_{\text{in}}$ to the work output $W_{\text{out}}$ and the change in internal energy:
$$Q_{\text{in}} = W_{\text{out}} + \Delta U$$If the engine operates in a steady cyclic state, $\Delta U = 0$ and the efficiency is
$$\eta = \frac{W_{\text{out}}}{Q_{\text{in}}}$$Question: A mass $m = 0.5\;\text{kg}$ attached to a spring of constant $k = 200\;\text{N m}^{-1}$ oscillates with amplitude $A = 0.10\;\text{m}$ in air. The damping force is $F_{\text{damp}} = -0.05 v$. Determine the rate at which mechanical energy is converted to heat after one quarter of a period.
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