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The Timpson Lyceum · University Physics I (2025)

Potential Energy and Energy Conservation – Lecture 7 (University Physics I)

Overview

This lecture introduces potential energy and the conservation of mechanical energy, building directly on the work–energy theorem from the previous chapter. It defines what makes a force conservative, derives the potential-energy functions for the two most important conservative forces — gravity and springs — and assembles the master energy-balance equation that handles non-conservative forces (like friction) as well. It closes with the deeper idea that total energy is always conserved (lost mechanical energy reappears as internal/thermal energy) and the calculus relationship that the force is the negative derivative of the potential-energy function. Mastering this material lets you solve a huge class of motion problems by bookkeeping energy at two points instead of tracking forces and accelerations the whole way.

Key Concepts

Conservative forces. A force is conservative if it has all four of these (equivalent) properties:

Gravity and ideal springs are conservative; friction is non-conservative (slide a box out and back and friction does negative work both ways — it never returns the energy, it dumps it as heat).

Work, defined. Work done by a force along a path from A to B: W=∫ABF⃗⋅ds⃗W = \int_A^B \vec{F}\cdot d\vec{s} Use the net force for total work, or a single force for the work done by that force alone.

Work–energy theorem. The total work done on an object equals its change in kinetic energy: Wtotal=12mvB2−12mvA2=KB−KAW_{\text{total}} = \tfrac{1}{2}mv_B^2 - \tfrac{1}{2}mv_A^2 = K_B - K_A with K=12mv2K = \tfrac{1}{2}mv^2.

Gravitational potential energy. Computing the work done by gravity along any path gives Wgrav=UA−UBW_{\text{grav}} = U_A - U_B, where U=mgyU = mgy (yy = height; the textbook often writes hh). Note the order: the work equals UA−UB=−ΔUU_A - U_B = -\Delta U, the negative change in potential energy — when UU drops, KK rises.

Elastic (spring) potential energy. A spring obeys Hooke's law, F=−kxF = -kx, where xx is the displacement from the spring's unstretched/equilibrium position (not where it started). Integrating the spring force gives U=12kx2U = \tfrac{1}{2}kx^2

Mechanical energy and its conservation. Mechanical energy is E=U+KE = U + K. If only conservative forces act, mechanical energy is conserved: U1+K1=U2+K2U_1 + K_1 = U_2 + K_2 This works for several conservative forces at once — just sum all the potential energies (gravity + spring + …).

The master equation (with non-conservative forces). In general, account for any other work done on the system: U1+K1+W=U2+K2U_1 + K_1 + W = U_2 + K_2 Here WW is the work done by non-conservative/external forces — positive (a push, a batted ball) or negative (friction, air resistance). For friction, W=−fk dW = -f_k \, d, using the distance traveled dd (not displacement), since friction always opposes motion along the whole path.

Conservation of total energy. "Lost" mechanical energy is never destroyed — it becomes internal energy (atoms jiggling faster = higher temperature). Across any process ΔU+ΔK+Δ(internal)=0\Delta U + \Delta K + \Delta(\text{internal}) = 0.

Force as the derivative of potential energy. Since UU is the negative integral of the force, the force is the negative derivative of UU. In 1-D: Fx=−dUdxF_x = -\dfrac{dU}{dx}. In 3-D, use partial derivatives (differentiate with respect to one variable while treating the others as constants): F⃗=(−∂U∂x, −∂U∂y, −∂U∂z)\vec{F} = \left(-\frac{\partial U}{\partial x},\ -\frac{\partial U}{\partial y},\ -\frac{\partial U}{\partial z}\right) It is usually far easier to differentiate a single scalar function UU to get the force than to integrate a vector force to get UU. Check: U=mgyU = mgy gives F⃗=(0,−mg,0)\vec{F} = (0, -mg, 0); U=12kx2U = \tfrac{1}{2}kx^2 gives Fx=−kxF_x = -kx.

Worked Example

Marble in a vertical loop. From what minimum height hh must a frictionless marble be released so it just completes a circular loop of radius RR? At the top of the loop (point B, height 2R2R) the minimum condition is that gravity alone supplies the centripetal acceleration: vB2R=g\dfrac{v_B^2}{R} = g, so vB2=gRv_B^2 = gR.

Apply energy conservation from release (at rest, height hh) to the top of the loop, with gravity the only force:

mgh=mg(2R)+12mvB2mgh=mg(2R)+12m(gR)h=2R+12R=52R\begin{aligned} mgh &= mg(2R) + \tfrac{1}{2}mv_B^2 \\ mgh &= mg(2R) + \tfrac{1}{2}m(gR) \\ h &= 2R + \tfrac{1}{2}R = \tfrac{5}{2}R \end{aligned}

The mass and gg cancel, so the required release height is 52R\tfrac{5}{2}R, independent of mass.

Skateboarder down a quarter-circle ramp (radius R=3 mR = 3\text{ m}, mass 25 kg25\text{ kg}, frictionless), starting from rest. Find the speed at the bottom, with g=9.8 m/s2g = 9.8\ \text{m/s}^2:

mgR=12mv2v=2gR=2(9.8)(3)≈7.67 m/s\begin{aligned} mgR &= \tfrac{1}{2}mv^2 \\ v &= \sqrt{2gR} = \sqrt{2(9.8)(3)} \approx 7.67\ \text{m/s} \end{aligned}

At the bottom the normal force must also supply the centripetal acceleration, giving N=mg+mv2R=3mg=3(25)(9.8)=735 NN = mg + \dfrac{mv^2}{R} = 3mg = 3(25)(9.8) = 735\ \text{N} — three times his weight, independent of RR.

Summary