Work-energy theorem — derivation from Newton's second law: Exercises

Before the numbers, one picture of what "work" and " $Δ K$ " mean so no symbol is unearned. In the figure below, look at the red arrow: that is the force $F$ doing work as the block slides through the displacement $d$ (the black double-arrow). The labels above show the block's kinetic energy at the start ( $K_{i}$ , on the left) and at the end ( $K_{f}$ , on the right). The whole page is just the statement written in red across the top of that picture: the work the red arrow does equals the jump from $K_{i}$ to $K_{f}$ .

Figure — Work-energy theorem — derivation from Newton's second law

Definition The symbols you need on every problem

$m$ = mass = how much "stuff" the object is, measured in kilograms ( $kg$ ). Bigger $m$ means harder to speed up.
$v$ = speed = how fast the object moves, in metres per second ( $m/s$ ). We write $v_{i}$ for the initial speed and $v_{f}$ for the final speed.
$x$ = position = where the object is along its line of travel, in metres ( $m$ ). When a force changes as the object moves, we write it as a function $F (x)$ — "the force at position $x$ ".
$d$ = displacement (distance) = how far the object moved, in metres ( $m$ ). It is the length the force acts through. (When $x$ runs from a start to an end, $d$ is just that total gap.)
$W$ = work = force multiplied by the distance $d$ it pushes through along the motion. Units: joule $J$ . If force and motion point the same way, $W > 0$ (speeds up); opposite way, $W < 0$ (slows down).
$K = \frac{1}{2} m v^{2}$ = kinetic energy = "energy of motion", built from the mass $m$ and speed $v$ above. Depends on speed squared, so it is always $\geq 0$ and never cares about direction. See Kinetic Energy.
$Δ K = K_{f} - K_{i}$ = final minus initial. Positive means the object sped up.
$F_{n e t}$ = net force = the single force you get after adding up every force with its sign (push forward is $+$ , friction or opposition is $-$ ). The theorem always uses this net force, never one force alone. See Newton's Second Law.
$θ$ = angle = the tilt (in degrees) between the force's direction and the direction of motion. When $θ = 0$ the force pulls straight along the path; larger $θ$ means more of the force is "wasted" sideways.
$t$ = time = how long something lasts, in seconds ( $s$ ).
$a$ = acceleration = how fast the speed changes each second, in $m/s^{2}$ ; from Newton's Second Law, $a = F_{n e t} / m$ .
$p$ = momentum = mass times velocity, $p = m v$ , in $kg\cdotm/s$ . Its change $Δ p$ is the "kick" a force delivers over time — the time-sibling of work. See Impulse–Momentum Theorem.

Level 1 — Recognition

Goal: identify which quantity the theorem hands you, with no traps.

Recall Solution L1.1

WHAT: we want $W_{n e t}$ . WHY the theorem: it says net work equals $Δ K$ directly — no forces, no distances needed once we know the energies. $W_{n e t} = Δ K = K_{f} - K_{i} = 30 - 12 = 18 J$ The $4 m$ is a distractor — the theorem already absorbed the distance into $K$ .

Recall Solution L1.2

WHY $W = F d$ here: the force is constant and parallel to motion, the only case where the integral $\int F d x$ collapses to a simple product. $W = F d = 8 \times 3 = 24 J \Rightarrow Δ K = 24 J$

Level 2 — Application

Goal: solve for a speed using the theorem, one force at a time.

Recall Solution L2.1

WHY $W = F d$ here: the force is constant and points along the motion, so the work integral collapses to a plain product (same reasoning as L1.2). $W_{n e t} = F d = 10 \times 5 = 50 J$ . Since $K_{i} = \frac{1}{2} (2) (0)^{2} = 0$ : $\frac{1}{2} m v_{f}^{2} = 50 \Rightarrow v_{f}^{2} = \frac{2 \times 50}{2} = 50 \Rightarrow v_{f} = 50 \approx 7.07 m/s$ WHY this beats $F = ma$ : we never solved for acceleration or time — force-over-distance went straight to speed.

Recall Solution L2.2

$K_{i} = \frac{1}{2} (1000) (20)^{2} = 200000 J$ , and $K_{f} = 0$ . $W_{n e t} = Δ K = 0 - 200000 = - 200000 J$ WHY negative: friction points opposite to the motion, so it removes kinetic energy. See Newton's Second Law for why that opposing force decelerates the car.

Recall Solution L2.3

$W_{n e t} = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = \frac{1}{2} (0.5) (1 0^{2} - 4^{2}) = 0.25 \times (100 - 16) = 0.25 \times 84 = 21 J$ Note we needed both speeds — only the change in $K$ matters, not either value alone.

Level 3 — Analysis

Goal: variable forces (integrate), and multiple forces (net first).

Recall Solution L3.1

WHY integrate: $F$ changes with position $x$ , so $W = F d$ is illegal; work is the area under the $F$ – $x$ graph (the red region in the figure), which is exactly $\int F d x$ . See Work done by a variable force. $W = \int_{0}^{2} 6 x d x = [3 x^{2}]_{0}^{2} = 3 (4) - 0 = 12 J$ $\frac{1}{2} (3) v_{f}^{2} = 12 \Rightarrow v_{f}^{2} = 8 \Rightarrow v_{f} = 22 \approx 2.83 m/s$

Recall Solution L3.2

WHY net first: the theorem uses the net force $F_{n e t}$ defined in the panel — add each force with its sign (forward push $+ 30$ , opposing friction $- 10$ ). $F_{n e t} = 30 - 10 = 20 N, W_{n e t} = 20 \times 5 = 100 J$ $\frac{1}{2} (4) v_{f}^{2} = 100 \Rightarrow v_{f}^{2} = 50 \Rightarrow v_{f} = 50 \approx 7.07 m/s$ Equivalently: your work $30 \times 5 = 150 J$ plus friction's work $- 10 \times 5 = - 50 J$ gives $100 J$ . Same answer.

Recall Solution L3.3

WHY $cos θ$ : only the component of force along the displacement does work — this is the dot product $F \cdot d r = F d cos θ$ , where $θ$ is the angle (defined in the panel) between the force and the motion. The vertical component does zero work because there is no vertical motion. $W = F d cos θ = 20 \times 6 \times 0.5 = 60 J$ $\frac{1}{2} (5) v_{f}^{2} = 60 \Rightarrow v_{f}^{2} = 24 \Rightarrow v_{f} = 24 = 26 \approx 4.90 m/s$

Level 4 — Synthesis

Goal: chain the theorem with gravity, inclines, and energy conservation.

The figure below shows the ramp for L4.1 and L4.2. The red block sits at the top; the vertical height it drops is marked $h$ . Notice the block travels along the slope (the black arrow), but the height label $h$ is the vertical side of the triangle — that distinction is the whole point of the next two solutions.

Recall Solution L4.1

WHY only gravity's vertical drop matters: in the figure, the normal force is perpendicular to the slope-arrow (does zero work), and gravity's work is $m g h$ regardless of ramp angle — it depends only on the vertical side $h$ marked in the picture, not the slope length. $W_{n e t} = m g h = 2 \times 10 \times 3 = 60 J$ $\frac{1}{2} (2) v_{f}^{2} = 60 \Rightarrow v_{f}^{2} = 60 \Rightarrow v_{f} = 60 \approx 7.75 m/s$ This is exactly Conservation of Mechanical Energy in disguise: gravity's work became kinetic energy.

Recall Solution L4.2

Gravity still does $m g h = 60 J$ (the vertical $h$ in the figure). Friction, however, acts along the slope arrow of length $5 m$ , so its negative work uses that slope distance: $W_{f r i c} = - 4 \times 5 = - 20 J, W_{n e t} = 60 - 20 = 40 J$ $\frac{1}{2} (2) v_{f}^{2} = 40 \Rightarrow v_{f}^{2} = 40 \Rightarrow v_{f} = 40 \approx 6.32 m/s$ Slower than the frictionless $7.75 m/s$ — friction stole $20 J$ .

Recall Solution L4.3

$W = \int_{0}^{3} 4 x d x = [2 x^{2}]_{0}^{3} = 18 J$ $\frac{1}{2} (1) v_{f}^{2} = 18 \Rightarrow v_{f}^{2} = 36 \Rightarrow v_{f} = 6 m/s$ Coasting means no further net work, so the speed stays $6 m/s$ .

Level 5 — Mastery

Goal: combine with power, momentum, and reason about limiting/edge cases.

Recall Solution L5.1

WHY power now: power is the rate of doing work, $P = \frac{W}{t}$ on average, using the time $t$ defined in the panel. See Power. $P_{a v g} = \frac{W}{t} = \frac{18}{2} = 9 W$ The theorem gave the work; power just divides by time.

Recall Solution L5.2

$K$ at $6 m/s$ : $\frac{1}{2} (3) (6)^{2} = 54 J$ . Direction does not matter — $K$ uses speed squared.

(a) Stopping: $W_{a} = 0 - 54 = - 54 J$ .
(b) Re-speeding: $W_{b} = 54 - 0 = + 54 J$ .
(c) Whole trip: start speed $6$ , end speed $6$ , so $Δ K = 54 - 54 = 0 J$ . Insight: the object ends moving the other way yet net work is zero — because Kinetic Energy can't tell left from right.

Recall Solution L5.3

Time route (impulse): $\int F d t = Δ p$ , the Impulse–Momentum Theorem, where $p = m v$ is the momentum defined in the panel. $F t = m v_{f} \Rightarrow v_{f} = \frac{12 \times 2}{3} = 8 m/s$ $Δ K = \frac{1}{2} (3) (8)^{2} - 0 = 96 J$ . Distance route (work): with acceleration $a = F / m = 4 m/s^{2}$ (the $a$ defined in the panel), we need the distance $d$ covered. WHY $d = \frac{1}{2} a t^{2}$ : starting from rest under constant acceleration, distance is the area under a straight velocity–time line — a triangle of base $t$ and height $a t$ , giving area $\frac{1}{2} (t) (a t) = \frac{1}{2} a t^{2}$ . This is the standard constant-acceleration kinematic result: $d = \frac{1}{2} a t^{2} = \frac{1}{2} (4) (4) = 8 m$ $W = F d = 12 \times 8 = 96 J = Δ K ✓$ Both siblings agree: time-integral gives momentum, distance-integral gives energy, and here they land consistently.

Recall Solution L5.4

Constant velocity ⇒ zero acceleration $a = 0$ ⇒ zero net force ⇒ friction $= 50 N$ opposing.

(a) $W_{a ppl i e d} = 50 \times 10 = + 500 J$ .
(b) $W_{f r i c} = - 50 \times 10 = - 500 J$ .
(c) $W_{n e t} = 500 - 500 = 0 J$ .
(d) $Δ K = 0 J$ — consistent, since the speed never changed. Edge lesson: you can pour in $500 J$ of effort and change the kinetic energy by nothing. The theorem only ever tracks the net.

Recall checkpoints

Recall Answer key (all values)

L1.1 $18 J$ · L1.2 $24 J$ · L2.1 $7.07 m/s$ · L2.2 $- 200000 J$ · L2.3 $21 J$ · L3.1 $12 J, 2.83 m/s$ · L3.2 $7.07 m/s$ · L3.3 $60 J, 4.90 m/s$ · L4.1 $7.75 m/s$ · L4.2 $6.32 m/s$ · L4.3 $6 m/s$ · L5.1 $9 W$ · L5.2 $- 54, + 54, 0 J$ · L5.3 $8 m/s, 96 J$ · L5.4 $+ 500, - 500, 0, 0$ .

Connections

Parent topic — the derivation these exercises drill.
Newton's Second Law · Kinetic Energy · Work done by a variable force · Conservation of Mechanical Energy · Power · Impulse–Momentum Theorem.