foundation26 min studyLesson 4 in Physics

Work, energy, and changing the question

Predict changes in speed from energy transfers while stating where friction, potentials, and system boundaries enter.

Before you begin

On this page

The question for today

What you’ll understand

Calculate work using the force component along a displacement.
Use a potential only under its stated conservative-force assumptions.
Balance mechanical energy with explicitly modeled transfers.

Sometimes you need a final speed without the complete trajectory. Energy methods change the question from how acceleration varies at every moment to how motion and stored energy change between states. They do not replace Newton's laws. They reorganize the same dynamics into a scalar account that becomes especially useful for many-particle systems and fields.

Work selects the part of a force along motion

A force can act without transferring kinetic energy. A table supports a sliding puck, but its vertical normal force is perpendicular to the puck's horizontal displacement. By contrast, horizontal friction removes kinetic energy. The relevant object is the dot product of force and displacement, accumulated along the actual path; force magnitude times path length is only a special case.

Positive net work increases kinetic energy; negative net work decreases it. Individual forces can do positive and negative work that cancel. A constant-speed lift provides a useful example: the lifting force adds energy while gravity removes the same amount from the object's kinetic-energy account. The object still gains gravitational potential energy when the object–Earth system is considered.

W_{\mathrm{net}}=\int_{\mathcal C}\mathbf F_{\mathrm{net}}\cdot d\mathbf r=\Delta K,\qquad K=\frac12mv^2

For a constant-mass Newtonian particle in an inertial frame, net work equals kinetic-energy change. Work and energy are measured in joules, with 1 J = 1 N·m. The path integral handles variable forces and curved paths.↔ Scroll wide equations horizontally.

A potential packages a conservative interaction

A conservative force does work that depends only on the endpoints, so we can assign a potential energy whose decrease equals that work. Near Earth's surface, choosing upward height gives gravitational potential proportional to height. An ideal spring stores energy quadratically in extension. The reference zero is arbitrary; only potential differences enter an energy balance.

Not every force admits a single-valued position potential. Sliding friction removes mechanical energy around a closed path, so it cannot be represented by simply assigning one potential value to each position. Velocity-dependent forces require additional care too. The useful question is whether a proposed potential reproduces the interaction under the assumptions, not whether the formula resembles familiar stored energy.

F_x=-\frac{dU}{dx},\qquad U_g=mgy,\qquad U_s=\frac12kx^2

The one-dimensional derivative relation holds for a conservative force. The displayed potentials assume uniform g and an ideal Hooke-law spring with stiffness k in N/m. Each potential may include an arbitrary additive constant.↔ Scroll wide equations horizontally.

Conserve the right account

For a particle subject only to conservative forces, kinetic plus potential energy is constant. If a nonconservative force does work, include that transfer explicitly. Be careful about system boundaries: friction between a sliding block and floor converts ordered motion into internal energy of both bodies. Mechanical energy falls, but total energy has not disappeared from the larger account.

Energy methods can identify inaccessible positions because kinetic energy cannot be negative. At a turning point in one dimension, kinetic energy momentarily vanishes while the slope of the potential can still imply a nonzero force. This potential-landscape viewpoint will return in quantum mechanics, where energy still constrains possibilities but does not specify a classical path between definite positions.

\Delta(K+U)=W_{\mathrm{nc}}

For a particle with conservative interactions represented by U, Wnc is the work of the remaining forces. This is an effective mechanical account, not a claim that all thermal or microscopic energy is already included in K + U.↔ Scroll wide equations horizontally.

Work it through

A rough ramp changes the answer

A $2.0\,\mathrm{kg}$ block starts from rest and descends a ramp through a vertical drop of $1.0\,\mathrm{m}$ . Sliding friction does $-4.0\,\mathrm{J}$ of work. Use $g=10\,\mathrm{m\,s^{-2}}$ . Find the final speed.

Set gravitational potential to zero at the bottom. The initial energy is $U_i=mgh=20\,\mathrm{J}$ and $K_i=0$ . The ramp's normal force does no work if it stays perpendicular to motion.
Account for friction instead of silently conserving mechanical energy: $K_f+0-(0+20)=-4.0\,\mathrm{J}$ . Therefore $K_f=16\,\mathrm{J}$ , smaller than the frictionless value.
Use $K_f=mv_f^2/2$ to obtain $v_f=\sqrt{2(16)/2.0}=4.0\,\mathrm{m\,s^{-1}}$ . The square root returns speed; the ramp geometry determines the velocity direction.

The result

The final speed is $4.0\,\mathrm{m\,s^{-1}}$ . The missing $4.0\,\mathrm{J}$ of mechanical energy is accounted for as internal energy when block and ramp are included.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

Watch motion trade with storage

Change damping ratio ζ and natural frequency ω₀ in a unit-mass oscillator released at x = 1 with zero velocity. Set damping to zero to compare turning points and equilibrium crossings. Nonzero damping removes mechanical energy; this is not the ramp's friction model.

Displacement x(t)Current time

Displacement x(t): 1
Velocity v(t): 0
Regime: underdamped

Damping ratio ζ0.2Natural frequency ω₀1.4

t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAt a turning point, zero speed need not mean zero force or zero stored energy.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The mass momentarily stops at a turning point, so its kinetic energy is zero. The spring is maximally deformed and stores the system's mechanical energy. Its restoring force is nonzero and reverses the motion.

Your turn

Try it yourself

An ideal spring with $k=200\,\mathrm{N\,m^{-1}}$ is compressed $0.10\,\mathrm{m}$ against a $0.50\,\mathrm{kg}$ block on a frictionless track. Starting from rest, what speed does the block have when the spring reaches its natural length?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compare initial spring potential with final kinetic energy; use metres for compression.

Compare with the solution

The spring initially stores $U_i=kx^2/2=1.0\,\mathrm{J}$ . With no dissipative work, that becomes $K_f=1.0\,\mathrm{J}$ . Thus $v=\sqrt{2K_f/m}=2.0\,\mathrm{m\,s^{-1}}$ . This result assumes negligible spring mass and an ideal force law over the full compression.

Keep the idea

Take it with you

Work accumulates the force component along the actual displacement.
A conservative potential describes an interaction, and its additive zero is arbitrary.
Mechanical energy conservation requires explicit assumptions about other transfers.

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