Classical-Mechanics · Unit 14 · Video 5 · Interactive Practice

Same Work, Different Rates: Power and the Flow of Kinetic Energy

IKey Formulas

Formula	Name	What you need
$P_{\text{ave}} = \dfrac{\Delta W}{\Delta t} = F_x\, v_{\text{ave},x}$	Average power	Work done over an interval $\Delta t$
$P = \lim\limits_{\Delta t \to 0} \dfrac{\Delta W}{\Delta t} = F_x\, v_x$	Instantaneous power	Force component along the motion and the velocity right now
$1\ \text{W} = 1\ \text{J}\cdot\text{s}^{-1}$	Watt	A rate — $1\ \text{kW}\cdot\text{h}$ is an amount of energy
$\dfrac{dK}{dt} = F_x\, v_x = P$	Work–energy theorem, differential form	Net force component and the velocity

Key Insight: A constant force need not deliver constant power — as $v_x$ grows, so does $P$ . Power is signed: a driving force feeds energy in, friction bleeds it out, and the signed sum of all powers is $dK/dt$ .

IIFalling Ball — Power Climbs as the Speed Climbs

Gravity is constant, yet it pumps energy in faster and faster: $P(t) = mg^2t$ rises straight from zero.

Time t 0.71 s

💡 Nothing here is special to gravity: for any constant force starting from rest, $P(t) = 2P_{\text{ave}}(0 \to t)$ , because a straight line through the origin averages half its final height.

IIIThe Sign of Power — Only the Component Along the Motion Counts

A force at angle $\theta$ to the motion delivers $P = Fv\cos\theta$ : energy in, energy out, or nothing at all.

Force angle θ 30°

💡 In the video's cup, kinetic friction is only $f_k = \mu_k N = 0.096\ \text{N}$ , so its power is $-0.14\ \text{W}$ — the same negative sign as a backward push, a far smaller drain.

IVPushing the Cup — Every Force Has Its Own Power

Each force on the cup delivers its own average power; the signed sum is the rate its kinetic energy grows.

Step 1

Step 1 — Work of the push

$F_x = F\cos 30\degree = (2.0)(0.8660) = 1.732\ \text{N}$

$W^{a} = F_x d = (1.732)(0.50) = 0.8660\ \text{J}$

Step 2 — Work of friction

$N = mg - F\sin 30\degree = 1.96 - 1.00 = 0.96\ \text{N}$

$f_k = \mu_k N = (0.1)(0.96) = 0.096\ \text{N}$

$W^{f} = -f_k d = -(0.096)(0.50) = -0.048\ \text{J}$

Step 3 — How long the push lasts

$W_{\text{net}} = 0.8660 - 0.048 = 0.818\ \text{J} = \tfrac{1}{2}mv_f^2$

$v_f = 2.860\ \text{m/s} \Rightarrow v_{\text{ave}} = \tfrac{1}{2}v_f = 1.430\ \text{m/s}$

$\Delta t = d / v_{\text{ave}} = 0.50/1.430 = 0.3496\ \text{s}$

Step 4 — The two powers and their sum

$P^{a}_{\text{ave}} = 0.8660/0.3496 = 2.48\ \text{W}$

$P^{f}_{\text{ave}} = -0.048/0.3496 = -0.14\ \text{W}$

$P^{a}_{\text{ave}} + P^{f}_{\text{ave}} = 2.34\ \text{W} = \dfrac{W_{\text{net}}}{\Delta t} = \dfrac{\Delta K}{\Delta t}$

VQuiz Questions

Problem 1 · Watts from Work and Time

Given: An elevator motor does $2.4 \times 10^{4}\ \text{J}$ of work on the cab during a $40\ \text{s}$ ride — find the average power delivered.

✅ Correct!

600\ \text{J}

of work is delivered every second, so

P_{\text{ave}} = 600\ \text{W}

❌ Right number, wrong quantity. The watt is a rate:

600\ \text{J/s} = 600\ \text{W}

. Joules would name an amount of energy, not how fast it is delivered.

❌ Not quite. Average power divides work by time:

P_{\text{ave}} = \Delta W / \Delta t

, not

\Delta W \times \Delta t

\Delta t / \Delta W

Show solution

Average power is work over the interval it took:

P_{\text{ave}} = \frac{\Delta W}{\Delta t} = \frac{2.4 \times 10^{4}\ \text{J}}{40\ \text{s}} = 6.0 \times 10^{2}\ \text{W}

Units confirm the arithmetic: $\text{J}/\text{s} = \text{W}$ . A $600\ \text{W}$ rating says how fast the motor delivers energy, never how much it holds — for the amount, multiply back: $(600\ \text{W})(40\ \text{s}) = 2.4 \times 10^{4}\ \text{J}$ .

Problem 2 · Signs and Perpendicular Forces

Given: A crate slides to the right at $v_x = 3.0\ \text{m/s}$ across a level floor. Kinetic friction of magnitude $4.0\ \text{N}$ acts backward, and the normal force is $30\ \text{N}$ — find the instantaneous power of friction and of the normal force.

✅ Correct! Friction opposes the motion, so it drains

12\ \text{J}

every second; the normal force is perpendicular to

\vec{v}

and delivers nothing.

❌ Check the sign. The friction component along the motion is

F_x = -4.0\ \text{N}

while

v_x = +3.0\ \text{m/s}

, so

P = F_x v_x

is negative — friction takes energy out.

❌ The normal force does no work.

P = F_x v_x

uses only the component along the motion. The normal force is vertical while the velocity is horizontal, so its component along the motion is zero.

❌ Not quite. Instantaneous power multiplies force by velocity,

P = F_x v_x

— it never divides them.

Show solution

Friction. Its component along the motion points backward: $F_x = -4.0\ \text{N}$ .

P_f = F_x v_x = (-4.0\ \text{N})(3.0\ \text{m/s}) = -12\ \text{W}

Normal force. It is perpendicular to the velocity, so its component along the motion is zero:

P_N = (0)(3.0\ \text{m/s}) = 0

Power is a signed scalar: a driving force gives $P > 0$ (energy in), friction gives $P < 0$ (energy out), and a perpendicular force gives $P = 0$ .

Problem 3 · Instantaneous vs Average (Falling Stone)

Given: A $0.50\ \text{kg}$ stone is released from rest and falls freely for $2.0\ \text{s}$ ( $g = 9.8\ \text{m/s}^2$ , no air resistance) — find the power delivered by gravity at $t = 2.0\ \text{s}$ and its average over the whole fall.

Instantaneous power at $t = 2.0\ \text{s}$ ?

Average power over $0 \to 2.0\ \text{s}$ ?

✅ Correct!

P(t) = mg^2t

rises linearly, so the value at the end is exactly twice the average since release.

❌ That is the average, not the instant.

P(t) = mg^2t

is the value right now; the average over

0 \to t

is half of it.

❌ Check the instantaneous power. Use

P = F_y v_y = (-mg)(-gt) = mg^2t

with

m = 0.50

g = 9.8

t = 2.0

❌ Check the average. The stone falls

h = \tfrac{1}{2}gt^2 = 19.6\ \text{m}

, so

W = mgh

and

P_{\text{ave}} = W/t

Show solution

Instantaneous. Falling from rest, $v_y = -gt = -19.6\ \text{m/s}$ and $F_y = -mg = -4.9\ \text{N}$ :

P = F_y v_y = (-4.9)(-19.6) = 96\ \text{W} \qquad \big(= mg^2t = (0.50)(9.8)^2(2.0)\big)

Average. The drop is $h = \tfrac{1}{2}gt^2 = \tfrac{1}{2}(9.8)(2.0)^2 = 19.6\ \text{m}$ , so gravity does $W = mgh = (0.50)(9.8)(19.6) = 96\ \text{J}$ :

P_{\text{ave}} = \frac{\Delta W}{\Delta t} = \frac{96\ \text{J}}{2.0\ \text{s}} = 48\ \text{W}

The ratio is exactly $2$ , and not by luck: $P(t) = mg^2t$ is a straight line through the origin, whose average height over $[0, t]$ is half its final height — $P_{\text{ave}} = \tfrac{1}{2}mg^2t$ .

Problem 4 · Power as the Rate of Change of $K$

Given: A $0.20\ \text{kg}$ cup is pushed along a table by a horizontal force of $1.5\ \text{N}$ while kinetic friction of $0.30\ \text{N}$ opposes the motion — find how fast its kinetic energy is changing at the instant it moves at $2.0\ \text{m/s}$ , and again at $4.0\ \text{m/s}$ .

$dK/dt$ at $v_x = 2.0\ \text{m/s}$ ?

$dK/dt$ at $v_x = 4.0\ \text{m/s}$ , same two forces?

✅ Correct! The net power is

F_{\text{net},x}v_x

with

F_{\text{net},x} = 1.2\ \text{N}

fixed, so doubling the speed doubles the rate at which

K

climbs.

❌ $K$ scales as $v^2$ , but its rate does not.

dK/dt = F_x v_x

is linear in

v_x

: doubling

v_x

doubles

dK/dt

, from

2.4\ \text{W}

4.8\ \text{W}

❌ Use the net force. Each force has its own power (

+3.0\ \text{W}

and

-0.60\ \text{W}

); their signed sum is the rate

K

changes.

❌ Check the dependence on $v_x$ .

dK/dt = F_{\text{net},x}v_x

— the forces are unchanged, so the rate is proportional to the speed.

Show solution

At $v_x = 2.0\ \text{m/s}$ , take the power of each force and add:

P_{\text{push}} = (1.5)(2.0) = 3.0\ \text{W}, \qquad P_{\text{fric}} = (-0.30)(2.0) = -0.60\ \text{W}

\frac{dK}{dt} = P_{\text{push}} + P_{\text{fric}} = 3.0 - 0.60 = 2.4\ \text{W}

Equivalently, use the net force directly: $F_{\text{net},x} = 1.5 - 0.30 = 1.2\ \text{N}$ , so $dK/dt = (1.2)(2.0) = 2.4\ \text{W}$ .

At $v_x = 4.0\ \text{m/s}$ , the forces are the same, so

\frac{dK}{dt} = (1.2\ \text{N})(4.0\ \text{m/s}) = 4.8\ \text{W}

This is the differential form of the work–energy theorem: $\dfrac{dK}{dt} = \dfrac{d}{dt}\big(\tfrac{1}{2}mv_x^2\big) = mv_x\dfrac{dv_x}{dt} = ma_xv_x = F_xv_x = P$ . Whatever net power flows in right now is exactly how fast $K$ is climbing right now.

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