Fiveable

⚾️Honors Physics Unit 9 Review

QR code for Honors Physics practice questions

9.1 Work, Power, and the Work–Energy Theorem

9.1 Work, Power, and the Work–Energy Theorem

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
⚾️Honors Physics
Unit & Topic Study Guides
Pep mascot

Work, Energy, and Power

Work, energy, and power describe how forces cause motion and transfer energy between objects. The work-energy theorem ties these ideas together by linking the net work done on an object directly to its change in kinetic energy. This section covers how to calculate work and power, distinguish kinetic from potential energy, and apply the work-energy theorem to solve problems.

Pep mascot
more resources to help you study

Work, Energy, and Power

Application of work-energy theorem, Work,Energy and Power – Physical sciences break 1.0

Application of the Work-Energy Theorem

The work-energy theorem says that the net work done on an object equals the change in its kinetic energy:

Wnet=ΔKE=12mvf212mvi2W_{net} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

  • WnetW_{net} = net work done on the object (in joules, J)
  • mm = mass of the object (kg)
  • viv_i = initial velocity (m/s)
  • vfv_f = final velocity (m/s)

This theorem follows directly from Newton's Second Law. A net force accelerates an object, and that acceleration over some displacement changes the object's speed. The work-energy theorem packages that relationship into a single equation.

Work itself is defined as the product of force and displacement in the direction of the force:

W=FdcosθW = Fd\cos\theta

  • FF = magnitude of the applied force
  • dd = magnitude of the displacement
  • θ\theta = angle between the force and displacement vectors

When θ=0°\theta = 0°, the force is fully along the displacement and cosθ=1\cos\theta = 1, so all the force contributes to work. When θ=90°\theta = 90°, the force is perpendicular to the displacement and does zero work (think of a centripetal force on an object in circular motion).

Steps for solving work-energy theorem problems:

  1. Identify the object and its initial and final velocities.

  2. Calculate the change in kinetic energy: ΔKE=12mvf212mvi2\Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2.

  3. Set Wnet=ΔKEW_{net} = \Delta KE to find the net work.

  4. If the problem asks for a force or displacement, use W=FdcosθW = Fd\cos\theta and solve for the unknown.

Example: A 2.0 kg box initially at rest is pushed across a frictionless surface by a 10 N horizontal force over 5.0 m. The work done is W=(10)(5.0)cos0°=50 JW = (10)(5.0)\cos 0° = 50 \text{ J}. By the work-energy theorem, 50=12(2.0)vf2050 = \frac{1}{2}(2.0)v_f^2 - 0, giving vf7.07 m/sv_f \approx 7.07 \text{ m/s}.

Application of work-energy theorem, Work-Energy Theorem | Boundless Physics

Calculation of Work and Power

Work uses the same formula from above:

W=FdcosθW = Fd\cos\theta

A few key angle cases to remember:

  • θ=0°\theta = 0°: Force is in the direction of motion → positive work (speeds the object up).
  • θ=180°\theta = 180°: Force opposes motion → negative work (slows the object down, like friction).
  • θ=90°\theta = 90°: Force is perpendicular → zero work (like a normal force on a flat surface).

Power measures how quickly work is done or energy is transferred. Its SI unit is the watt (W), where 1 W = 1 J/s.

P=WtP = \frac{W}{t}

There's also an instantaneous form that's useful when you know force and velocity directly:

P=FvcosθP = Fv\cos\theta

  • PP = power (watts)
  • WW = work (joules)
  • tt = time (seconds)
  • FF = force, vv = velocity, θ\theta = angle between them

Steps for calculating power:

  1. Find the work done (using W=FdcosθW = Fd\cos\theta) or identify the force and velocity.
  2. Choose the appropriate formula: use P=W/tP = W/t if you know total work and time, or P=FvcosθP = Fv\cos\theta if you know force and velocity.
  3. Solve and check units.

Example: A motor lifts a 50 kg crate 10 m in 5.0 s at constant velocity. The work done against gravity is W=mgh=(50)(9.8)(10)=4900 JW = mgh = (50)(9.8)(10) = 4900 \text{ J}. The power output is P=4900/5.0=980 WP = 4900/5.0 = 980 \text{ W}.

Kinetic vs. Potential Energy

Kinetic energy is the energy an object has because of its motion:

KE=12mv2KE = \frac{1}{2}mv^2

It's always zero or positive (mass and v2v^2 can't be negative). A 1,000 kg car traveling at 20 m/s has KE=12(1000)(20)2=200,000 JKE = \frac{1}{2}(1000)(20)^2 = 200{,}000 \text{ J}. Notice that doubling the speed quadruples the kinetic energy, which is why high-speed collisions are so much more destructive.

Potential energy is stored energy due to an object's position or configuration:

  • Gravitational PE: PEg=mghPE_g = mgh, where hh is the height above a chosen reference point.
  • Elastic PE: PEe=12kx2PE_e = \frac{1}{2}kx^2, where kk is the spring constant and xx is the displacement from the spring's equilibrium position.

The total mechanical energy of a system is the sum of kinetic and potential energy:

Etotal=KE+PEE_{total} = KE + PE

When only conservative forces (gravity, springs) do work, total mechanical energy is conserved:

ΔKE+ΔPE=0\Delta KE + \Delta PE = 0

This means any kinetic energy gained comes from potential energy lost, and vice versa. A ball dropped from a height converts gravitational PE into KE as it falls.

When non-conservative forces like friction are present, they dissipate mechanical energy as thermal energy (heat). In that case, total mechanical energy decreases:

Wfriction=ΔKE+ΔPEW_{friction} = \Delta KE + \Delta PE

The work done by friction is negative, so it reduces the system's mechanical energy.

Momentum and Impulse

While the work-energy theorem connects force and displacement to energy changes, momentum and impulse connect force and time to changes in motion.

Momentum is the product of mass and velocity:

p=mv\vec{p} = m\vec{v}

It's a vector quantity, so direction matters. A 0.15 kg baseball moving at 40 m/s has a momentum of 6.0 kg·m/s in the direction of travel.

Impulse is the change in an object's momentum, and it equals the net force multiplied by the time interval over which it acts:

J=Δp=FnetΔt\vec{J} = \Delta\vec{p} = \vec{F}_{net}\Delta t

This is sometimes called the impulse-momentum theorem, and it's structurally parallel to the work-energy theorem:

TheoremConnectsEquation
Work-EnergyForce × displacement → change in KEWnet=ΔKEW_{net} = \Delta KE
Impulse-MomentumForce × time → change in momentumJ=Δp\vec{J} = \Delta\vec{p}
Just as energy is conserved in the absence of non-conservative forces, momentum is conserved when no net external force acts on a system. Both conservation laws are powerful tools for analyzing collisions and interactions.