Back to Course Overview

Module 3: Energy and Work

 

PHYS-2325 M3L1 Work and Kinetic Energy

"Energy cannot be created or destroyed; it can only be changed from one form to another."
— Albert Einstein


Newton's laws told you how forces change motion. Energy methods give you a powerful alternative: instead of tracking forces and accelerations instant by instant, you can relate the net work done on an object directly to its change in kinetic energy — the Work-Kinetic Energy Theorem. This single equation, derived using calculus from Newton's second law, often solves problems in one step that would otherwise require multiple force equations and kinematics. You will learn to calculate work done by constant forces, by gravity, by spring forces, and by any variable force using integration — then connect work to power, the rate at which energy is transferred.

A graph of force versus displacement with the area under the curve shaded, representing the work done by a variable force.
Work done by a variable force equals the area under the F vs. x curve — a definite integral.

Course Competencies and Learning Objectives

A ★ indicates that this page contains content related to that LO.

CC3.1 Analyze the energy transformations associated with the work-energy theorem

★ LO3.1.1 Define kinetic energy and calculate it for an object of known mass and velocity

★ LO3.1.2 Calculate the work done by a constant force using W = F · d = Fd cos φ

★ LO3.1.3 Apply the Work-Kinetic Energy Theorem to find the speed of an object after a net work is done on it

★ LO3.1.4 Calculate the work done by the gravitational force on objects moving along any path

★ LO3.1.5 Calculate the work done by a spring force using Hooke's Law and integration

★ LO3.1.6 Calculate the work done by a variable force using graphical integration and functional integration (definite integral)

★ LO3.1.7 Define power and calculate average and instantaneous power using P = dW/dt = F · v

Required Reading

Click the blue buttons to go to the OpenStax reading assignments. Complete all six sections before watching the videos.

7.1 Kinetic Energy 7.2 Work and Kinetic Energy 7.3 Work Done by the Gravitational Force 7.4 Work Done by a Spring Force 7.5 Work Done by a General Variable Force 7.6 Power

 

Optional Reading

Explore More

Want additional practice with work, energy, and power? These free resources complement the reading and offer interactive tools.

Media

Work through the pre-lecture videos for each reading section first, then watch the mini-lectures for detailed worked examples.

Video 1: Pre-Lecture — 7.1 Kinetic Energy

Pre-Lecture: Kinetic Energy

This video introduces kinetic energy as the energy an object possesses because of its motion. Using calculus, we derive that kinetic energy K = ½mv2 and establish its units (joules, J). The video emphasizes that kinetic energy depends on speed squared — doubling speed quadruples kinetic energy.

  • Kinetic energy: K = ½mv2 — scalar quantity, always ≥ 0, SI unit: joule (J = kg·m2/s2)
  • K depends on the magnitude of velocity (speed), not direction
  • Relative kinetic energy: K depends on the reference frame — always state the frame
  • Connecting to Newton's second law: the calculus derivation that leads to the work-energy theorem

Time: Video pending upload to Canvas

Video 2: Pre-Lecture — 7.2 Work and Kinetic Energy

Pre-Lecture: Work, the Dot Product, and the Work-Kinetic Energy Theorem

This video defines work done by a constant force as the dot product F · d = Fd cos φ, where φ is the angle between the force and displacement vectors. It then states and derives the Work-Kinetic Energy Theorem: Wnet = ΔK.

  • Work by a constant force: W = F · d = Fd cos φ — scalar; can be positive, negative, or zero
  • Dot product geometry: W = 0 when force is perpendicular to displacement (no energy transfer)
  • Work-Kinetic Energy Theorem: Wnet = Kf − Ki = ΔK
  • Derived from Newton's 2nd Law using Fnet = ma and kinematics (or the integral of F·dx)

Time: Video pending upload to Canvas

Video 3: Pre-Lecture — 7.3 Work Done by the Gravitational Force

Pre-Lecture: Gravitational Work — Lifting and Lowering

This video applies the work formula to the gravitational force. The key result: work done by gravity depends only on the vertical displacement (Δy), not on the path taken. It also introduces the work done against gravity by an applied force when lifting an object.

  • Work by gravity: Wg = mgΔy (positive going down, negative going up; take +y upward)
  • Path independence: gravity does the same work regardless of the route — this is a signature of conservative forces
  • Lifting at constant velocity: Wapplied = −Wg = mgh (applied force does positive work equal to the gain in height)
  • Lifting with acceleration: Wnet = ΔK ≠ 0 — the applied force exceeds mg during acceleration

Time: Video pending upload to Canvas

Video 4: Pre-Lecture — 7.4 Work Done by a Spring Force

Pre-Lecture: Hooke's Law and Spring Work via Integration

The spring force Fs = −kx varies with position, so the work it does must be calculated by integration. This video derives Ws = ½kxi2 − ½kxf2 and applies it to objects attached to springs.

  • Hooke's Law: Fs = −kx — restoring force, opposes displacement from equilibrium
  • Spring work (calculus): Ws = ∫xixf (−kx) dx = ½kxi2 − ½kxf2
  • From equilibrium (xi = 0): Ws = −½kxf2 (spring does negative work when stretched or compressed)
  • Applying Wnet = ΔK with spring work to find speed after compression/extension

Time: Video pending upload to Canvas

Video 5: Pre-Lecture — 7.5 Work Done by a General Variable Force

Pre-Lecture: Integration as the General Method for Variable Force Work

When the force varies arbitrarily with position, neither the constant-force formula nor Hooke's Law applies. This video shows how to calculate work as a definite integral — both graphically (area under the F vs. x curve) and analytically (evaluating ∫F(x) dx).

  • General definition: W = ∫xixf F(x) dx — the area under the F vs. x graph from xi to xf
  • Graphical integration: estimate W by counting grid squares or using geometric shapes
  • Functional integration: when F(x) is given as an equation, evaluate the antiderivative analytically
  • Three-dimensional case: W = ∫F · dr = ∫(Fxdx + Fydy + Fzdz)

Time: Video pending upload to Canvas

Video 6: Pre-Lecture — 7.6 Power

Pre-Lecture: Power — the Rate of Doing Work

Power is the rate at which work is done or energy is transferred. This video introduces average power and derives instantaneous power as the dot product of force and velocity, P = F · v. It connects power to the familiar units of watts and horsepower.

  • Average power: Pavg = ΔW / Δt — SI unit: watt (W = J/s)
  • Instantaneous power: P = dW/dt = F · v = Fv cos φ
  • Unit conversions: 1 hp = 746 W; 1 kWh = 3.6 × 106 J
  • Integrating power to get work (or energy): W = ∫ P dt — used when power varies with time

Time: Video pending upload to Canvas

Video 7: Mini-Lecture — Work and Kinetic Energy

Mini-Lecture: Applying the Work-Kinetic Energy Theorem

This video works a complete example applying Wnet = ΔK: given an object's initial speed, mass, and one or more forces acting over a displacement, find the final speed. The strategy is to calculate work done by each force, sum them for Wnet, then solve for vf.

  • Calculate work done by each force: W = Fd cos φ for each
  • Identify forces that do zero work: normal force, centripetal force (perpendicular to motion)
  • Sum all works: Wnet = W1 + W2 + ... = ΔK = ½mvf2 − ½mvi2
  • Solve for vf: vf = √(vi2 + 2Wnet/m)

Time: Video pending — to be recorded by instructor

Video 8: Mini-Lecture — Work Along a Ramp

Mini-Lecture: Work by Multiple Forces on an Inclined Surface

A block moves along a rough inclined ramp under an applied force. This video demonstrates how to calculate the work done by each force — applied force, gravity, normal force, and friction — using the dot product, then applies the Work-Kinetic Energy Theorem to find the final speed.

  • Gravity work along a ramp: Wg = −mgd sin θ (opposing upward displacement)
  • Normal force work: WN = 0 (perpendicular to displacement)
  • Friction work: Wf = −μkNd = −μkmg cos θ · d (always negative — opposes motion)
  • Net work gives ΔK: solve for vf from initial rest or given vi

Time: Video pending — to be recorded by instructor

Video 9: Mini-Lecture — Work by a Variable Force: Graphical and Functional Integration

Mini-Lecture: Two Methods for Finding Work from a Variable Force

This video solves the same problem two ways: first by estimating the area under an F vs. x graph (graphical integration), then by evaluating the antiderivative analytically (functional integration). The comparison shows when each method is appropriate.

  • Graphical method: divide the area under F(x) into rectangles or triangles; sum their areas; estimate uncertainty
  • Functional method: given F(x) = axn, integrate: W = ∫xixf axn dx = [axn+1/(n+1)]xixf
  • Both methods yield the same result — graphical is approximate, functional is exact
  • Use Wnet = ΔK to find the final speed after the variable force acts

Time: Video pending — to be recorded by instructor

Video 10: Mini-Lecture — Power

Mini-Lecture: Calculating Instantaneous and Average Power

This video works through power calculations in two contexts: finding the constant power output needed to accelerate an object over a distance, and integrating a time-varying power function to find total work (energy transferred). It reinforces the link P = F · v.

  • At constant speed: P = Fv (driving force equals friction; power goes to heat)
  • During acceleration: P = Fnetv = mav — power increases as speed increases
  • Time-varying power: W = ∫titf P(t) dt — integrate to find total energy delivered
  • Converting between watts, kilowatts, and horsepower; kWh as a unit of energy

Time: Video pending — to be recorded by instructor

Practice and Apply - Conceptual

Strategy: Applying the Work-Kinetic Energy Theorem

Arrange the following steps in the correct order for solving a work-kinetic energy problem.

Order Items

Arrange the following items in the correct order.

  1. Identify all forces acting on the object during the displacement.
  2. Calculate the work done by each force: W = Fd cos φ.
  3. Identify forces that do zero work (perpendicular to displacement).
  4. Sum all non-zero works to find Wnet.
  5. Apply Wnet = ΔK = ½mvf2 − ½mvi2.
  6. Solve for the unknown (usually vf or displacement d).
  7. Check: verify units are joules (J) and the answer is physically reasonable.

Which Forces Do Zero Work?

From the scenarios below, select all forces that do zero work on the moving object.

Select All

Select all forces that do zero work.

  • The normal force on a block sliding along a horizontal surface.
  • The tension in a string on a ball moving in a horizontal circle.
  • The centripetal force on an object in uniform circular motion.
  • The normal force on a block sliding up a ramp (force perpendicular to ramp surface).
  • The gravitational force on a ball thrown upward (does negative work).
  • The friction force on a sliding block (does negative work opposing motion).
  • An applied horizontal force on a block moving horizontally in the same direction.
  • The spring force on a mass being stretched away from equilibrium.

Sort by Sign of Work Done

Sort each scenario into the correct category based on the sign of the work done by the specified force.

Sort Items

Sort each item into the correct category.

Positive Work (W > 0)
Negative Work (W < 0)
Zero Work (W = 0)
  • Gravity on a ball falling downward.
  • Gravity on a ball thrown upward.
  • Applied push on a box sliding in the same direction as the push.
  • Kinetic friction on a sliding block.
  • Spring force on a mass being pulled away from equilibrium (spring pulls back).
  • Applied force pulling a mass toward a spring's equilibrium from extension.
  • Normal force on a block moving horizontally on a flat surface.
  • Air resistance on a car moving forward.

Key Concept Check

Click each card to flip it and test your understanding of work, kinetic energy, and power.

State the Work-Kinetic Energy Theorem. What does it mean physically?
Answer

Wnet = ΔK = Kf − Ki = ½mvf2 − ½mvi2. Physically: the net work done on an object equals the change in its kinetic energy. Positive net work speeds the object up; negative net work slows it down; zero net work means constant speed.

 

Why must work done by a spring be calculated by integration rather than W = Fd cosφ?
Answer

Because the spring force varies with position: Fs(x) = −kx. The formula W = Fd cosφ is only valid for a constant force. For a variable force, work must be computed as a definite integral: Ws = ∫xixf (−kx)dx = ½kxi2 − ½kxf2.

 

A car moves at constant velocity. Its engine exerts a forward force. Is the net work on the car zero? Explain.
Answer

Yes, the net work is zero. Constant velocity means ΔK = 0, so Wnet = 0 by the Work-Kinetic Energy Theorem. The engine does positive work, but friction (and drag) does equal negative work. The engine's power output is entirely dissipated as heat by friction — none goes into kinetic energy.

 

How does the work done by gravity depend on the path taken? What does this tell you about gravity as a force?
Answer

The work done by gravity depends only on the vertical displacement Δy — not on the path. Wg = mgΔy regardless of whether the object takes a straight line, a curved path, or a ramp. This path-independence is the defining property of a conservative force, which allows us to define a potential energy function (covered in Module 4).

 

What does the area under an F(x) vs. x graph represent? How do you calculate it when F(x) is a given function?
Answer

The area under the F(x) vs. x graph between xi and xf represents the work done by that force over that displacement. When F(x) is given as a function, calculate it as the definite integral: W = ∫xixf F(x) dx. Evaluate by finding the antiderivative F(x) → G(x), then compute G(xf) − G(xi).

 

Write the formula for instantaneous power in terms of force and velocity. When is this most useful?
Answer

P = dW/dt = F · v = Fv cos φ. This is most useful when you know the force and the velocity at a specific instant and need the power at that moment — for example, finding the power output of a motor driving a vehicle at a given speed, or integrating P(t) to find total work: W = ∫P dt.

 

Practice and Apply - Computational

Important: Key Equations and Strategy Reminders

Kinetic energy: K = ½mv2

Work by a constant force: W = Fd cos φ = F · d

Work-Kinetic Energy Theorem: Wnet = ΔK = ½mvf2 − ½mvi2

Work by gravity: Wg = −mgΔy  (positive downward; use Wg = mgh for downward drop h)

Work by a spring: Ws = ½kxi2 − ½kxf2

Work by a variable force: W = ∫xixf F(x) dx

Power: Pavg = W/Δt  |  P = dW/dt = F · v

  • Work is a scalar — add works algebraically (watch signs).
  • Normal force and centripetal force do zero work — perpendicular to displacement.
  • Report energy in joules (J), power in watts (W). Three significant figures unless stated otherwise.

Sample Problem Videos

Watch each sample problem video, then attempt the corresponding written practice problem below.

Sample Problem 1: Work Done by a Constant Force in Unit-Vector Notation

Sample Problem: W = F · d Using Component Form

A force given in unit-vector notation acts on an object undergoing a displacement also given in unit-vector notation. This video calculates the work using the dot product component method and verifies with the magnitude-angle formula.

  • Dot product: F · d = Fxdx + Fydy + Fzdz
  • Equivalent to: W = Fd cos φ where φ is the angle between F and d
  • Key check: confirm sign makes physical sense (positive work → force has a component along motion)

Time: Video pending — to be recorded by instructor

Sample Problem 2: Work and Kinetic Energy

Sample Problem: Finding Final Speed Using Wnet = ΔK

An object starts at rest (or a given speed) and has several forces acting on it over a displacement. This video finds the final speed by calculating each force's work, summing for Wnet, and applying the Work-Kinetic Energy Theorem.

  • Calculate W for each force; identify zero-work forces immediately
  • Wnet = W1 + W2 + ... (algebraic sum)
  • Solve Wnet = ½mvf2 − ½mvi2 for vf

Time: Video pending — to be recorded by instructor

Sample Problem 3: Work Done by Two Constant Forces — Industrial Spies

Sample Problem: Net Work and Final Speed from Two Applied Forces

Two forces act simultaneously on an object moving through a displacement. This video finds the net work done by both forces and uses the Work-Kinetic Energy Theorem to determine the object's final speed. It illustrates that only the component of each force along the displacement contributes to work.

  • W1 = F1d cos φ1; W2 = F2d cos φ2
  • Wnet = W1 + W2 = ΔK
  • Compare to alternative: find Fnet first, then compute Wnet = Fnetd cos φnet

Time: Video pending — to be recorded by instructor

Sample Problem 4: Work Done Along a Ramp

Sample Problem: Work by Applied Force, Gravity, Friction, and Normal on an Incline

A block is pushed up a rough ramp by an applied force. This video calculates the work done by each of the four forces (applied, gravity, friction, normal) individually, then sums them and applies Wnet = ΔK to find the final speed.

  • Applied force work: Wapp = Fappd cos φapp
  • Gravity work: Wg = −mgd sin θ (component of weight along ramp × distance)
  • Friction work: Wf = −μkmg cos θ · d (always negative, opposes motion)
  • Normal force work: WN = 0 (perpendicular to motion)

Time: Video pending — to be recorded by instructor

Sample Problem 5: Work Done on an Accelerating Elevator Cab

Sample Problem: Elevator Work — Cable Tension, Gravity, and ΔK

An elevator cab accelerates upward over a given displacement. This video finds the work done by the cable tension and by gravity, verifies that their sum equals ΔK, and finds the speed at the end of the acceleration phase.

  • Find tension using Newton's 2nd Law: T − mg = ma → T = m(g + a)
  • WT = Td (positive, upward force × upward displacement)
  • Wg = −mgd (negative, gravity opposes upward motion)
  • Wnet = WT + Wg = mad = ΔK ✓

Time: Video pending — to be recorded by instructor

Sample Problem 6: Work Done in Lifting — Paul Anderson

Sample Problem: Work Against Gravity — Weightlifting Application

A real-world application: calculating the work done lifting a heavy load through a vertical distance. This video emphasizes that work done against gravity equals mgh regardless of path, and connects this to energy requirements in athletic and engineering contexts.

  • Wapp = mgh (lifting at constant speed, applied force = weight)
  • If accelerating: Wapp = m(g + a)h (more work required)
  • Path independence: work done against gravity depends only on height, not horizontal displacement
  • Practical check: compare to energy content of food (1 kcal = 4186 J)

Time: Video pending — to be recorded by instructor

Sample Problem 7: Work Done by a Spring

Sample Problem: Spring Work via Integration — Stretch and Compression

A spring (spring constant k) is compressed or stretched from one position to another. This video applies the spring work integral Ws = ½kxi2 − ½kxf2 to two cases: compressing from equilibrium and further compressing from an already-compressed state.

  • From equilibrium (xi = 0): Ws = −½kxf2 (negative: spring opposes compression)
  • Between two non-zero positions: Ws = ½k(xi2 − xf2)
  • Sign interpretation: spring does negative work when moving away from equilibrium, positive work when returning toward it

Time: Video pending — to be recorded by instructor

Sample Problem 8: Work Done by Spring to Change Kinetic Energy

Sample Problem: Applying Wnet = ΔK with Spring Work

A block attached to a spring is released from rest at a compressed position. This video applies the Work-Kinetic Energy Theorem with spring work as the only force doing work (frictionless surface, normal force perpendicular) to find the block's speed when the spring returns to equilibrium.

  • At release (xi = −x0) to equilibrium (xf = 0): Ws = ½kx02 (positive — spring releases energy)
  • Wnet = Ws = ΔK = ½mvf2 − 0
  • vf = x0√(k/m) — this is the maximum speed for a spring-block system
  • Connection to simple harmonic motion: vmax = Aω (previewed here, detailed in Module 7)

Time: Video pending — to be recorded by instructor

Sample Problem 9: Work Calculated by Graphical Integration

Sample Problem: Estimating Work from an F vs. x Graph

Given a graph of a variable force F(x) versus position x, this video estimates the work done by counting grid squares, using geometric decomposition (rectangles + triangles), and comparing to the exact analytical result where possible.

  • Identify regions: divide the graph into simple geometric shapes
  • Area of each region = work done by that portion of the force profile
  • Watch signs: area above the x-axis is positive work; below is negative work
  • Total work = algebraic sum of all region areas

Time: Video pending — to be recorded by instructor

Sample Problem 10: Work, Two-Dimensional Integration

Sample Problem: Work by a 2D Variable Force Along a Curved Path

A force with both x and y components varies along a two-dimensional path. This video evaluates the work integral W = ∫F · dr = ∫(Fxdx + Fydy) by parameterizing the path and integrating each component separately.

  • Set up: W = ∫path (Fxdx + Fydy)
  • Parameterize the path: express y as a function of x (or use a parameter t)
  • Substitute and integrate each term independently
  • Check whether the result depends on the path chosen (non-conservative) or only on endpoints (conservative)

Time: Video pending — to be recorded by instructor

Sample Problem 11: Kinetic Energy — Train Crash

Sample Problem: Comparing Kinetic Energies — Mass vs. Speed

This problem compares the kinetic energies of two objects (e.g., a slow heavy train and a fast light car) to illustrate how K = ½mv2 depends more sensitively on speed than on mass. It also applies the Work-Kinetic Energy Theorem to find the braking force needed to stop each.

  • K = ½mv2: doubling mass doubles K; doubling speed quadruples K
  • Stopping: Wbraking = −ΔK = −Ki (bring to rest); W = −Fd → F = Ki/d
  • Compare stopping distances for different initial kinetic energies at the same braking force

Time: Video pending — to be recorded by instructor

Sample Problem 12: Power — Force and Velocity / Integrating Power to Get Speed

Sample Problem: Power Output at a Given Speed; Speed from Integrated Power (Funny Car)

Two power problems in one video: (1) finding the instantaneous power output of an engine given force and velocity using P = Fv; (2) integrating a time-dependent power function P(t) to find total work, then using Wnet = ΔK to find the final speed of a dragster ("funny car") from rest.

  • Part 1: P = Fv — given F and v at a specific instant, calculate instantaneous power
  • Part 2: W = ∫0t P(t') dt' — integrate P(t) to get total work delivered
  • Apply Wnet = ½mvf2 − 0 → vf = √(2W/m)
  • Real-world context: NHRA Top Fuel/Funny Car performance data

Time: Video pending — to be recorded by instructor

Practice Problem 1 — Work by Multiple Forces and Final Speed

A 4.00-kg block starts from rest on a frictionless horizontal surface. A constant force F1 = 12.0 N acts at 0° (along the direction of motion) and a second constant force F2 = 6.00 N acts at 60.0° above the direction of motion. Both forces act over a displacement d = 5.00 m. (a) Find the work done by each force. (b) Find the final speed of the block.

Show Solution

Given: m = 4.00 kg, vi = 0, F1 = 12.0 N at φ1 = 0°, F2 = 6.00 N at φ2 = 60.0°, d = 5.00 m, frictionless surface

Part (a) — Work by each force:

W1 = F1d cos φ1 = (12.0)(5.00) cos 0° = (12.0)(5.00)(1) = 60.0 J

W2 = F2d cos φ2 = (6.00)(5.00) cos 60.0° = (6.00)(5.00)(0.500) = 15.0 J

WN = 0 (normal force perpendicular to motion)

Wg = 0 (gravity perpendicular to horizontal motion)

Part (b) — Final speed:

Wnet = W1 + W2 = 60.0 + 15.0 = 75.0 J

Wnet = ½mvf2 − 0 → vf = √(2Wnet/m) = √(2 × 75.0 / 4.00)

vf = √(37.5) = 6.12 m/s

Practice Problem 2 — Spring Work and Final Speed

A 2.50-kg block rests on a frictionless surface and is connected to a spring with spring constant k = 800 N/m. The spring is compressed 0.120 m from equilibrium and released from rest. (a) Calculate the work done by the spring as the block moves from x = −0.120 m to x = 0. (b) Find the speed of the block at x = 0.

Show Solution

Given: m = 2.50 kg, k = 800 N/m, xi = −0.120 m, xf = 0, vi = 0

Part (a) — Spring work:

Ws = ½kxi2 − ½kxf2 = ½(800)(0.120)2 − ½(800)(0)2

Ws = ½(800)(0.0144) − 0 = 400 × 0.0144 = 5.76 J

Positive: the spring pushes the block toward equilibrium — force and displacement are in the same direction.

Part (b) — Speed at x = 0:

Wnet = Ws = 5.76 J (normal force and gravity do zero work on horizontal surface)

Wnet = ½mvf2 → vf = √(2Wnet/m) = √(2 × 5.76 / 2.50)

vf = √(4.608) = 2.15 m/s

Check using vmax = x0√(k/m) = 0.120√(800/2.50) = 0.120 × 17.9 = 2.15 m/s ✓

Practice Problem 3 — Work by a Variable Force via Integration

A variable force acts on a 3.00-kg object along the x-axis: F(x) = (3.00x2 − 2.00x) N, where x is in meters. The object starts at xi = 1.00 m with speed 2.00 m/s. (a) Calculate the work done by this force as the object moves from x = 1.00 m to x = 3.00 m. (b) Find the object's speed at x = 3.00 m.

Show Solution

Given: m = 3.00 kg, F(x) = (3.00x2 − 2.00x) N, xi = 1.00 m, xf = 3.00 m, vi = 2.00 m/s

Part (a) — Work by variable force:

W = ∫1.003.00 (3.00x2 − 2.00x) dx

= [x3 − x2]1.003.00

= [(3.00)3 − (3.00)2] − [(1.00)3 − (1.00)2]

= [27.0 − 9.00] − [1.00 − 1.00]

= 18.0 − 0 = 18.0 J

Part (b) — Speed at x = 3.00 m:

Ki = ½mvi2 = ½(3.00)(2.00)2 = 6.00 J

Wnet = ΔK = Kf − Ki → Kf = Ki + W = 6.00 + 18.0 = 24.0 J

Kf = ½mvf2 → vf = √(2Kf/m) = √(2 × 24.0 / 3.00) = √(16.0) = 4.00 m/s

Ready to Move On?

Lesson Checklist

Before moving on to the next lesson, confirm you can do each of the following:

  • ☐ Calculate kinetic energy K = ½mv2 for an object with given mass and speed.
  • ☐ Calculate work done by a constant force using W = Fd cos φ and identify when W = 0.
  • ☐ Apply the Work-Kinetic Energy Theorem Wnet = ΔK to find the final speed of an object.
  • ☐ Calculate work done by gravity: Wg = −mgΔy (path-independent).
  • ☐ Calculate work done by a spring force: Ws = ½kxi2 − ½kxf2.
  • ☐ Evaluate the work done by a variable force using a definite integral: W = ∫F(x) dx.
  • ☐ Estimate work graphically from an F vs. x graph by computing the area under the curve.
  • ☐ Calculate average power Pavg = W/Δt and instantaneous power P = F · v.
  • ☐ Integrate a power function P(t) to find total work: W = ∫P dt.

Work and kinetic energy are the entry point to all of energy physics. In the next lesson you will add potential energy and the full conservation of energy framework — the most powerful problem-solving tool in all of classical mechanics.