Classical-Mechanics · Unit 14 · Video 4 · Interactive Practice

Work as Area Under the Curve: The Work–Kinetic Energy Theorem

IKey Formulas

Formula	Name	What you need
$W = \lim\limits_{N \to \infty} \sum\limits_{j=1}^{N} (F_{x,j})_{\text{ave}}\, \Delta x_j = \int_{x_i}^{x_f} F_x(x)\, dx$	Work of a varying force	$F_x$ as a function of $x$
$W^{\text{spring}} = -\tfrac{1}{2} k \left( x_f^{\,2} - x_i^{\,2} \right)$	Work done by a spring	Hooke's law $F_x = -kx$
$W^{\text{total}} = K_f - K_i = \Delta K$	Work–kinetic energy theorem	The work of every force
$v_f = \sqrt{\,v_i^{\,2} + \dfrac{2 W^{\text{total}}}{m}\,}$	Final speed from work	$K = \tfrac{1}{2} m v^2$

Key Insight: The area under the $F_x$ -versus- $x$ curve is the change in kinetic energy — one curve, one area, one answer, with no equations of motion in sight.

IIChopping the Trip into Rectangles

No single force value fits the whole trip, but over one thin slice the force barely changes at all.

Slices N 8

IIIThe Spring's Signed Area

The spring's work is the signed area under $F_x = -kx$ between the two stretches.

Start position 3 m End position 1 m

💡 Released from rest at $x_i$ on a frictionless table, the block reaches the relaxed position with $\tfrac{1}{2} m v^2 = \tfrac{1}{2} k x_i^{\,2}$ — the spring's whole area converted into kinetic energy.

IVReading the Sign of the Work

Only the force component along the displacement feeds kinetic energy; a perpendicular push feeds none.

Force angle θ 60°

VQuiz Questions

Problem 1 · Area Under a Varying Force

Given: a bead on the $x$ -axis is pushed by the force $F_x = 3x$ (newtons, with $x$ in metres) from $x_i = 0$ to $x_f = 4\ \text{m}$ — find the work this force does.

✅ Correct! The region under

F_x = 3x

from

0

4\ \text{m}

is a triangle of area

\tfrac{1}{2}(4\ \text{m})(12\ \text{N}) = 24\ \text{J}

❌ That is a force, not a work.

12

is the value of

F_x

in newtons at the very end of the trip; it still has to be integrated over the

4\ \text{m}

❌ That is the "too big" rectangle.

F_{\max}\Delta x = (12\ \text{N})(4\ \text{m})

assumes the force held its final value the whole way, but it started at zero.

❌ That is the average force, not the work. Because

F_x

is linear, its average over the trip is

\bar{F} = \tfrac{0 + 12}{2} = 6\ \text{N}

— still multiply it by the

4\ \text{m}

to get the work.

❌ Not quite. Work is the area under the graph of

F_x

versus

x

, not a single force value.

Show solution

The force varies with position, so integrate it:

W = \int_{0}^{4} 3x\, dx = \left[ \tfrac{3}{2} x^2 \right]_{0}^{4} = \tfrac{3}{2}(16) = 24\ \text{J}

Check it geometrically. The graph is a straight line from $(0, 0\ \text{N})$ to $(4\ \text{m}, 12\ \text{N})$ , so the area is a triangle: $\tfrac{1}{2}(\text{base})(\text{height}) = \tfrac{1}{2}(4)(12) = 24\ \text{J}$ .

Equivalently, because the force is linear its average value is $\bar{F} = \tfrac{0 + 12}{2} = 6\ \text{N}$ , and $\bar{F}\,\Delta x = (6)(4) = 24\ \text{J}$ — the $6\ \text{N}$ is the average force, not the work.

Problem 2 · Work Done by a Spring

Given: a block on a frictionless table is attached to a spring of stiffness $k = 50\ \text{N/m}$ , with $x$ measured from the relaxed position. The block moves from $x_i = 0.40\ \text{m}$ to $x_f = 0.10\ \text{m}$ — find the work done by the spring.

✅ Correct! The spring goes from greater tension to lesser tension, so its force lines up with the motion and it pours energy into the block.

❌ Right size, wrong sign.

x_f^{\,2} - x_i^{\,2} = 0.01 - 0.16 = -0.15\ \text{m}^2

, and the leading minus in

-\tfrac{1}{2}k(x_f^{\,2} - x_i^{\,2})

flips that back to positive.

❌ You squared the displacement.

\tfrac{1}{2}k(x_i - x_f)^2

is not the work — each endpoint is squared separately, then subtracted.

❌ Check the subtraction.

0.01 + 0.16 = 0.17

adds the squares; the formula subtracts them:

0.01 - 0.16 = -0.15

❌ Not quite. Use

W = -\tfrac{1}{2}k\left(x_f^{\,2} - x_i^{\,2}\right)

with both stretches in metres.

Show solution

Hooke's law gives $F_x = -kx$ , so the work is the signed area under that line:

W = \int_{x_i}^{x_f} (-kx)\, dx = \left[ -\tfrac{1}{2}kx^2 \right]_{x_i}^{x_f} = -\tfrac{1}{2}k\left(x_f^{\,2} - x_i^{\,2}\right)

Substituting $k = 50\ \text{N/m}$ , $x_f = 0.10\ \text{m}$ , $x_i = 0.40\ \text{m}$ :

W = -\tfrac{1}{2}(50)\left(0.10^2 - 0.40^2\right) = -25\,(0.01 - 0.16) = -25(-0.15) = +3.75\ \text{J}

Why positive? Both endpoints enter as squares, so only the magnitudes matter: $|x_f| < |x_i|$ makes $x_f^{\,2} - x_i^{\,2}$ negative, and the leading minus makes $W > 0$ . A spring moving toward lesser tension always does positive work — stretch it further instead and $W$ turns negative.

Problem 3 · From Area to Final Speed

Given: a $1.0\ \text{kg}$ cart starts at rest at $x = 0$ on a frictionless horizontal track. The only horizontal force on it is $F_x = 4x$ (newtons, $x$ in metres), acting from $x = 0$ to $x = 3.0\ \text{m}$ — find its final speed.

✅ Correct! The area under

F_x = 4x

18\ \text{J}

, and starting from rest that whole area becomes

\tfrac{1}{2}mv_f^{\,2}

❌ You dropped the $\tfrac{1}{2}$ .

\sqrt{W/m} = \sqrt{18} = 4.24

; because

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

, the speed is

\sqrt{2W/m}

❌ That uses the end force for the whole trip.

(12\ \text{N})(3\ \text{m}) = 36\ \text{J}

overestimates: the force grows from zero, so the triangular area is only

18\ \text{J}

❌ That is $v_f^{\,2}$ , not $v_f$ .

2W/m = 36\ \text{m}^2/\text{s}^2

— take the square root.

❌ Not quite. Get

W

as the area under

F_x = 4x

, then set

W = \Delta K = \tfrac{1}{2}mv_f^{\,2}

Show solution

Step 1 — the work is the area:

W = \int_{0}^{3} 4x\, dx = \left[ 2x^2 \right]_{0}^{3} = 18\ \text{J}

(Geometric check: a triangle of base $3\ \text{m}$ and height $F_x(3) = 12\ \text{N}$ , area $\tfrac{1}{2}(3)(12) = 18\ \text{J}$ .)

Step 2 — apply the work–kinetic energy theorem. It is the only horizontal force and the cart starts at rest, so $K_i = 0$ and

W = \Delta K = \tfrac{1}{2}mv_f^{\,2} - 0 \quad \Longrightarrow \quad v_f = \sqrt{\frac{2W}{m}}

v_f = \sqrt{\frac{2(18\ \text{J})}{1.0\ \text{kg}}} = \sqrt{36} = 6.0\ \text{m/s}

No equation of motion was needed: one integral over position gave the speed directly.

Problem 4 · The Pushed Cup (Video Example)

Given: a $0.20\ \text{kg}$ cup, starting from rest, is pushed $\Delta x = 0.50\ \text{m}$ along a horizontal table by a $2.0\ \text{N}$ force aimed $30\degree$ above the horizontal. The coefficient of kinetic friction is $\mu_k = 0.10$ and $g = 9.8\ \text{m/s}^2$ .

What is the normal force on the cup?

What is the cup's final speed?

✅ Correct! The tilted push carries part of the cup's weight, so

N = 0.96\ \text{N}

, friction is only

0.096\ \text{N}

, and the total work

0.818\ \text{J}

becomes kinetic energy.

❌ That is the full weight $mg$ . The push has an upward component

F\sin 30\degree = 1.0\ \text{N}

, so the table only has to supply the rest.

❌ Check the direction of $F_y$ . The push tilts upward, so it relieves the table:

N = mg - F\sin 30\degree

, not

mg + F\sin 30\degree

❌ Not quite. The cup does not accelerate vertically:

N + F\sin 30\degree - mg = 0

❌ You rounded too early.

0.818\ \text{J}

rounded to

0.80\ \text{J}

gives

\sqrt{8.0} = 2.8\ \text{m/s}

; carry the unrounded work and round once, at the end.

❌ Only the horizontal component does work. Using the full

2.0\ \text{N}

instead of

F\cos 30\degree = 1.732\ \text{N}

inflates the work.

❌ Not quite. Total the work first,

W = (F\cos 30\degree - \mu_k N)\Delta x

, then use

v_f = \sqrt{2W/m}

Show solution

Step 1 — audit the forces. Gravity and the normal force are perpendicular to the motion, so each does zero work. Only the push and friction do work.

Step 2 — components of the push:

F_x^{\,a} = (2.0)\cos 30\degree = 1.732\ \text{N}, \qquad F_y^{\,a} = (2.0)\sin 30\degree = 1.0\ \text{N}

Step 3 — vertical balance gives the normal force:

N + F_y^{\,a} - mg = 0 \;\Longrightarrow\; N = 1.96 - 1.0 = 0.96\ \text{N}

The tilted push holds part of the cup up, so $N$ is noticeably less than the weight $mg = 1.96\ \text{N}$ .

Step 4 — friction and total work:

f_k = \mu_k N = (0.10)(0.96) = 0.096\ \text{N}

W = \left(F_x^{\,a} - f_k\right)\Delta x = (1.732 - 0.096)(0.50) = 0.818\ \text{J}

Step 5 — the theorem. The cup starts at rest, so $K_f = W = 0.818\ \text{J}$ and

v_f = \sqrt{\frac{2K_f}{m}} = \sqrt{\frac{2(0.818)}{0.20}} = \sqrt{8.18} = 2.9\ \text{m/s}

Round the work to $0.80\ \text{J}$ too early and the speed comes back $2.8\ \text{m/s}$ — carry unrounded values and round once, at the end.

Solved: 0 / 4