Classical-Mechanics · Unit 15 · Video 1 · Interactive Practice

The Scalar Product: Work as a Dot Product of Force and Displacement

IKey Formulas

Formula	Name	What you need
$\vec A \cdot \vec B = AB\cos\theta$	Scalar product (geometric)	Two magnitudes and the angle between them
$\vec A \cdot \vec B = A_xB_x + A_yB_y + A_zB_z$	Scalar product (components)	Cartesian components — no angle
$K = \tfrac{1}{2}m(\vec v \cdot \vec v)$	Kinetic energy	Mass and the velocity vector
$W = \vec F \cdot \Delta\vec r = (F\cos\beta)\,\Delta x$	Work of a constant force	Force and displacement vectors

Key Insight: The sign of a scalar product is the sign of $\cos\theta$ — positive when the two vectors cooperate, zero when they are perpendicular, negative when they oppose.

IIVisualization 1 — One Number, Two Routes

The same pair of vectors yields one number two ways: matching components summed, or $AB\cos\theta$ .

💡 Challenge: place the two tips so that $\vec A \cdot \vec B = 0$ while neither vector has zero length.

IIIVisualization 2 — Three Forces on the Incline

Three forces act on the sliding block; the dot product sorts their work into zero, negative, and positive.

Step 1

Step 1 — Coordinates. Origin at the top of the incline, $+x$ down the slope, $+y$ perpendicular to the surface, so the motion is one-dimensional:

\Delta\vec r = l\,\hat i = 3.0\,\hat i\ \text{m}

Step 2 — Normal force. The block never leaves the surface, so $a_y = 0$ :

N - mg\cos\theta = 0 \quad\Longrightarrow\quad N = mg\cos\theta = 33.9\ \text{N}

f_k = \mu_k N = (0.2)(33.9\ \text{N}) = 6.79\ \text{N}

Step 3 — Normal force. It is pure $\hat j$ ; the displacement is pure $\hat i$ :

W^N = (mg\cos\theta\,\hat j)\cdot(l\,\hat i) = mgl\cos\theta\,(\hat j\cdot\hat i) = 0

Step 4 — Friction. Pure $\hat i$ , pointing up the slope against $\Delta\vec r$ :

W^f = (-\mu_k mg\cos\theta\,\hat i)\cdot(l\,\hat i) = -\mu_k mgl\cos\theta

= -(0.2)(4.0\,\text{kg})(9.8\,\text{m/s}^2)(3.0\,\text{m})\cos 30\degree = -20.4\ \text{J}

Step 5 — Gravity. Its $\hat j$ part dies against the $\hat i$ displacement:

W^g = (mg\sin\theta\,\hat i - mg\cos\theta\,\hat j)\cdot(l\,\hat i) = mgl\sin\theta

= (4.0\,\text{kg})(9.8\,\text{m/s}^2)(3.0\,\text{m})\sin 30\degree = 58.8\ \text{J}

Step 6 — Total work.

W = 0 - 20.4\ \text{J} + 58.8\ \text{J} = +38.4\ \text{J} = mgl(\sin\theta - \mu_k\cos\theta)

IVVisualization 3 — Net Work Becomes Kinetic Energy

Every joule of net work on the block reappears as kinetic energy $K = \tfrac{1}{2}m(\vec v \cdot \vec v)$ .

Distance slid (m) 0.00

VQuiz Questions

Problem 1 · Dot Product from Components

Given: $\vec A = 3\hat i + 4\hat j$ and $\vec B = 2\hat i - \hat j$ — find $\vec A \cdot \vec B$ .

✅ Correct!

(3)(2) + (4)(-1) = 6 - 4 = 2

— matching components multiplied, then added.

❌ Close, but check the signs.

B_y = -1

, so the second term is

(4)(-1) = -4

, not

+4

❌ Not quite. The scalar product returns a single number, not a vector — you multiplied components without summing them.

❌ Not quite.

11.2 = |\vec A|\,|\vec B|

, which equals the dot product only when

\theta = 0\degree

Show solution

Use the component form — no angle needed:

\vec A \cdot \vec B = A_xB_x + A_yB_y = (3)(2) + (4)(-1) = 6 - 4 = 2

Cross-check with the geometric form. Here $|\vec A| = 5$ and $|\vec B| = \sqrt{5} \approx 2.24$ , so

\cos\theta = \frac{\vec A \cdot \vec B}{|\vec A||\vec B|} = \frac{2}{11.18} = 0.179 \quad\Longrightarrow\quad \theta = 79.7\degree

Just under a right angle, so a small positive number is exactly what to expect.

Problem 2 · Sign of the Work

Given: a constant force of magnitude $F = 50\ \text{N}$ acts at $\beta = 120\degree$ to a displacement of magnitude $|\Delta\vec r| = 4.0\ \text{m}$ — find the work $W$ .

✅ Correct!

\cos 120\degree = -\tfrac{1}{2}

, so

W = (50)(4.0)(-0.5) = -100\ \text{J}

: the force drains energy.

❌ Close, but check the sign. Beyond

90\degree

the cosine is negative:

\cos 120\degree = -\tfrac{1}{2}

, not

+\tfrac{1}{2}

❌ Not quite.

173 = (50)(4.0)\sin 120\degree

— work uses the cosine, the component along the displacement.

❌ Not quite. Work vanishes only at exactly

\beta = 90\degree

; at

120\degree

the force still has a component along

\Delta\vec r

Show solution

Work is the scalar product of force and displacement:

W = \vec F \cdot \Delta\vec r = F\,|\Delta\vec r|\cos\beta = (50\ \text{N})(4.0\ \text{m})\cos 120\degree

W = (200)(-0.500) = -100\ \text{J}

The component of the force along the motion is $F\cos\beta = -25\ \text{N}$ — it points backwards along $\Delta\vec r$ , so the work is negative.

Problem 3 · Incline with New Numbers

Given: a block of mass $m = 2.0\ \text{kg}$ slides $l = 4.0\ \text{m}$ down a $\theta = 30\degree$ incline with $\mu_k = 0.50$ , taking $g = 9.8\ \text{m/s}^2$ .

What work does gravity do?

What is the total work on the block?

✅ Correct!

W^N = 0

W^f = -33.9\ \text{J}

W^g = +39.2\ \text{J}

, so

\Sigma W = +5.3\ \text{J}

— barely positive, because

\mu_k\cos\theta

nearly cancels

\sin\theta

❌ Check gravity's work. Only the component along

\Delta\vec r

survives the dot product, and that component is

mg\sin\theta

❌ Check the total. Friction's work is negative —

W^f = -\mu_k mgl\cos\theta = -33.9\ \text{J}

— and the normal force contributes nothing.

Show solution

Step 1 — Normal force (from $a_y = 0$ ):

N = mg\cos\theta = (2.0)(9.8)\cos 30\degree = 17.0\ \text{N}, \qquad W^N = 0

Step 2 — Friction (opposes $\Delta\vec r$ ):

W^f = -\mu_k N l = -(0.50)(17.0\ \text{N})(4.0\ \text{m}) = -33.9\ \text{J}

Step 3 — Gravity (only the downhill component):

W^g = mgl\sin\theta = (2.0)(9.8)(4.0)(0.500) = 39.2\ \text{J}

Step 4 — Total:

\Sigma W = 0 - 33.9 + 39.2 = +5.3\ \text{J} = mgl(\sin\theta - \mu_k\cos\theta)

Since $\mu_k = 0.50$ and $\cos 30\degree = 0.866$ , the bracket is $0.500 - 0.433 = 0.067$ — the block still speeds up, but only just.

Problem 4 · From Work to Speed (Video Example)

Given: in the video's incline the total work on the $m = 4.0\ \text{kg}$ block is $+38.4\ \text{J}$ and it starts from rest — find its speed at the bottom using $K = \tfrac{1}{2}m(\vec v \cdot \vec v)$ .

✅ Correct!

38.4 = \tfrac{1}{2}(4.0)v^2

gives

v^2 = 19.2\ \text{m}^2/\text{s}^2

and

v = 4.4\ \text{m/s}

❌ Not quite. Don't drop the

\tfrac{1}{2}

K = \tfrac{1}{2}mv^2

, so

v^2 = 2K/m = 19.2

, not

K/m = 9.6

❌ Not quite.

\sqrt{38.4} = 6.2

ignores the mass entirely — solve

38.4 = \tfrac{1}{2}(4.0)v^2

instead.

❌ Close, but that is $v^2$ .

19.2

has units

\text{m}^2/\text{s}^2

; take the square root to get the speed.

Show solution

Starting from rest, all of the net work appears as kinetic energy:

K = \tfrac{1}{2}m(\vec v \cdot \vec v) = \tfrac{1}{2}mv^2 = 38.4\ \text{J}

v^2 = \frac{2K}{m} = \frac{2(38.4)}{4.0} = 19.2\ \text{m}^2/\text{s}^2 \quad\Longrightarrow\quad v = 4.4\ \text{m/s}

The normal force added nothing and friction removed $20.4\ \text{J}$ ; only the surviving $+38.4\ \text{J}$ shows up as speed.

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