1 2 3 4 5 6 7

0:007 lessons across the whole video1:10:15

You will be able to

Compute the work done by a constant force using the dot product and the displacement vector
Evaluate work done by a variable force by integrating along a path
Apply the work-energy theorem to relate changes in kinetic energy to net work
Define potential energy for conservative forces such as gravity and springs
Use conservation of mechanical energy to solve motion problems with turning points and speeds
Account for non-conservative forces such as friction in an energy balance

Good to know first

Newton's second law and free-body diagrams
One-dimensional and two-dimensional kinematics with vectors
Basic integration and differentiation
Familiarity with the dot product of vectors

Syllabus

7 lessons and a final project

1
Setting up motion with Newton's second law
0:00–7:007 min of video
This lesson shows how to set up the motion of an object moving in a circle by drawing its forces and writing Newton's second law for the radial direction. It prepares the ground for the energy ideas that follow in the chapter.
2 formulas3 exercises3 questions
2
Work by a Constant Force and the Dot Product
7:00–16:4010 min of video
Work is the force multiplied by the displacement, and when the force and displacement point in different directions we use the dot product to take only the part of the force along the motion. Starting from kinematics with a constant force, we derive that the change in kinetic energy equals the work done.
3 formulas3 exercises3 questions
3
Work by a variable force along a path
16:40–28:2012 min of video
When the force changes with position, the work is no longer simply F times d. We split the path into tiny steps, add up the small works, and the sum becomes an integral, which is the area under the force-versus-position graph.
3 formulas3 exercises3 questions
4
Kinetic energy and the work-energy theorem
28:20–38:2010 min of video
The work done by the net force on a particle equals its change in kinetic energy. Applied to a spring-mass system, this gives conservation of mechanical energy, E = ½kA² = ½mv² + ½kx².
4 formulas3 exercises3 questions
5
Conservative Forces and Potential Energy
38:20–48:2010 min of video
A conservative force is one whose work depends only on the start and end positions, not the path taken, which lets us define a potential energy U(x). Total mechanical energy (kinetic plus potential) then stays constant when only conservative forces act.
3 formulas4 exercises3 questions
6
Conservation of Mechanical Energy in Worked Examples
48:20–58:2010 min of video
When only conservative forces (like springs and gravity) do work, the total mechanical energy K + U stays constant. Friction is a non-conservative force that removes mechanical energy, and its work must be added to the energy balance.
3 exercises3 questions
7
Small changes, Taylor expansion, and why the binomial/exponential approximations matter
58:20–1:10:1512 min of video
This section builds the mathematical tool of linearising a function: a small change in F is approximately the slope times the small change in x. It then uses Taylor expansions to approximate functions such as (1+x)^n and e^x when x is small.
4 formulas4 exercises3 questions

Final project

Energy Audit of a Spring-Block System with Friction

You will analyse a block on a horizontal spring, first without friction and then with kinetic friction, using only the work-energy theorem and conservation of energy. You will confirm the analytic results with a small Python calculation and use a binomial approximation to check a small-friction result.

Free preview · Lesson 1 in full

Setting up motion with Newton's second law

This lesson shows how to set up the motion of an object moving in a circle by drawing its forces and writing Newton's second law for the radial direction. It prepares the ground for the energy ideas that follow in the chapter.

0:00 to 7:00 in the video

Scratchpad

Draw and write next to the video.

The idea

The chapter opens with a problem set on the board: a small object swings on a string in a vertical circle, and the lecturer asks us to work out how its motion behaves. The first step in any mechanics problem like this is to draw the object and every force acting on it, then choose a direction along which to apply Newton's second law. The board shows the object at the top of the circle with two forces: the weight, labelled mg, pointing straight down, and the tension T in the string, pointing toward the centre of the circle.

Newton's second law says that the net force on an object equals its mass times its acceleration, $\sum F = m a$ . Forces are vectors, so we apply the law separately in each direction. For motion in a circle, the useful direction is the radial one, pointing from the object toward the centre. The acceleration in that direction is the centripetal acceleration, $v^{2} / R$ , where v is the speed and R is the radius of the circle.

Applying this at the top of the circle, both T and mg point toward the centre, so the net radial force is T + mg. The board writes exactly this: $T + m g = m a = m \frac{v^{2}}{R}$ . This single equation links the tension, the weight, the speed and the radius. It is the starting point of the problem.

The lecturer then moves the object to other positions on the circle and repeats the process, which is the key habit: the direction of the forces changes with position, so you redraw the forces each time. Only after setting up this equation does the lecture bring in energy, which will make it easier to find the speed without solving the force equation directly.

Key points

The problem is a ball on a string moving in a vertical circle; first draw the forces on the object. 0:45
At the top of the circle, tension and weight both point toward the centre, so the radial equation is T + mg = mv²/R. 3:30
The direction of the forces changes around the circle, so the diagram must be redrawn at each position. 5:00
Writing Newton's second law gives a relation between forces, speed and radius, which energy methods will later simplify. 6:30

Formulas

\sum F = m a

The net force on an object equals its mass times its acceleration.

$\sum F$: sum (net) of all forces
$m$: mass of the object
$a$: acceleration

T + m g = m \frac{v^{2}}{R}

At the top of the circle, tension plus weight equals mass times speed squared over radius.

$T$: tension in the string
$m g$: weight of the object
$m$: mass
$v$: speed of the object
$R$: radius of the circle

Watch out

Do not invent an extra force such as a separate 'centripetal force'. Only the weight and the tension act. Together they give the net force toward the centre. 3:30
Do not reuse one free-body diagram at every point on the circle; the directions of T and mg relative to the centre change. 5:00
Work in the direction of the acceleration (toward the centre), not in an arbitrary direction. 2:31

Practice

Calculate Solution checked

A 0.5 kg ball on a string of radius 1.0 m moves in a circle at the top of its path with speed 4.0 m/s. Taking g = 10 m/s², find the tension T in the string at the top, using $T + m g = m \frac{v^{2}}{R}$ .

Hint

Solve for T: T = mv²/R − mg.

Solution

Compute the centripetal term: m v²/R = 0.5 × 4.0² / 1.0 = 0.5 × 16 / 1.0 = 8.0 N.
Compute the weight: mg = 0.5 × 10 = 5.0 N.
Subtract: T = 8.0 − 5.0 = 3.0 N.

3.0 N

Explain Solution checked

At the top of the circle, why do we write T + mg and not T − mg in Newton's second law for the radial direction?

Hint

Choose the direction toward the centre as positive. Which way does each force point at the top?

Solution

The radial direction points from the object toward the centre, which is downward at the top.
The weight mg points down, so it points toward the centre and counts as positive.
The string pulls the object toward the centre too, so T is also positive.
Both forces are positive, so the net radial force is T + mg.

Both the tension and the weight point toward the centre at the top, so they add: T + mg.

Calculate Solution checked

A ball on a string moves in a vertical circle of radius 2.5 m. At the top the string just goes slack, so T = 0. Taking g = 10 m/s², find the speed v of the ball at the top, using $T + m g = m \frac{v^{2}}{R}$ .

Hint

Put T = 0. The mass m cancels on both sides.

Solution

With T = 0 the equation becomes mg = m v²/R.
Cancel m: g = v²/R, so v² = gR = 10 × 2.5 = 25.
Take the square root: v = 5 m/s.

5 m/s

On your scratchpad

Sketch a vertical circle with the ball at the top. Draw arrows for the weight (down) and tension (toward the centre), then write the radial equation T + mg = mv²/R next to it.

Quiz

What is the first step in setting up this kind of problem?
1. Solve for the speed directly
2. Draw the object and all forces acting on it
3. Measure the radius of the circle
4. Assume the tension is zero
Show the answer
B. Draw the object and all forces acting on it · The lecturer starts by drawing the object and the forces (weight and tension) before applying Newton's second law.
0:45 watch the moment
At the top of the circle, which way do the tension and the weight point?
1. Both point away from the centre
2. Both point toward the centre
3. Tension points up, weight points sideways
4. Both point up
Show the answer
B. Both point toward the centre · At the top, the string pulls the object toward the centre and gravity also pulls it down toward the centre.
3:30 watch the moment
What does Newton's second law give for the radial direction at the top?
1. T - mg = mv/R
2. T + mg = mv squared over R
3. T = mg
4. mg = 0
Show the answer
B. T + mg = mv squared over R · Both forces point toward the centre, so the net radial force T + mg equals m v² / R.
5:00 watch the moment

Open lesson 1, with the video →

Made by Beyond from the video “5. Work-Energy Theorem and Law of Conservation of Energy” by YaleCourses. The video plays in YouTube's own player. Watch it on YouTube · All lessons

7 lessons and a final project

Setting up motion with Newton's second law

Work by a Constant Force and the Dot Product

Work by a variable force along a path

Kinetic energy and the work-energy theorem

Conservative Forces and Potential Energy

Conservation of Mechanical Energy in Worked Examples

Small changes, Taylor expansion, and why the binomial/exponential approximations matter

Energy Audit of a Spring-Block System with Friction

Setting up motion with Newton's second law

The idea

Key points

Formulas

Watch out

Practice

On your scratchpad

Quiz