A Pushed Body Gains Kinetic Energy

solved KineticEnergy

Q. A body of mass 2 kg $m = 2$ kg initially at rest $v_{0} = 0$ m/s moves under the action of an applied horizontal force $F_{h} = | F_{h} |$ of 7 N $| F_{h} | = 7$ N on a table with coefficient of kinetic friction = 0.1 $F_{t} = −0.1 \cdot m \cdot g$ . Taking g as 9.8 m/s² $g = 9.8$ m/s², compute the change in kinetic energy of the body in 10 s $Δ t = 10$ s, in joules.

Inputs

mass

m

2 kg

force

| F_{h} |

7 N

gravity

g

9.8 m/s²

time

Δ t

10.00 s

Outputs

gain

Δ K

How it's solved

Solved with KineticEnergy

Given

$m = 2$ kg
$| F_{h} | = 7$ N
$g = 9.8$ m/s²
$Δ t = 10$ s
$v_{0} = 0$ m/s from “initially at rest”
$F_{h} = | F_{h} |$ from “applied horizontal force”
$F_{t} = −0.1 \cdot m \cdot g$ from “coefficient of kinetic friction = 0.1”

Assumed

$x_{0} = 0$ m

Find $Δ K$ .

Working

applied horizontal force $F_{h} = | F_{h} |$
$F_{h} = 7$
$F_{h} = 7$ N
coefficient of kinetic friction = 0.1 $F_{t} = −0.1 \cdot m \cdot g$
$F_{t} = −0.1 \cdot 2 \cdot 9.8$
$F_{t} = −1.96$ N
net force is the resultant(body, sliding, x) $F_{net} = F_{h} + F_{t} + 0$
$F_{net} = 7 + (−1.96) + 0$
$F_{net} = 5.04$ N
second law(sliding, x) $F_{net} = m \cdot a$
$5.04 = 2 \cdot a$
$a = 2.52$ m/s²
velocity after a time(sliding, x) $v_{1} = v_{0} + a \cdot Δ t$
$v_{1} = 0 + 2.52 \cdot 10$
$v_{1} = 25.2$ m/s
kinetic energy of a body(body, sliding, 2) $K_{1} = \frac{m \cdot {v_{1}}^{2}}{2}$
$K_{1} = \frac{2 \cdot {25.2}^{2}}{2}$
$K_{1} = 635.04$ J
kinetic energy of a body(body, sliding, 1) $K_{0} = \frac{m \cdot {v_{0}}^{2}}{2}$
$K_{0} = \frac{2 \cdot 0^{2}}{2}$
$K_{0} = 0$ J
kinetic energy gained(body, sliding) $Δ K = K_{1} - K_{0}$
$Δ K = 635.04 - 0$
$Δ K = 635$ J
$Δ K \approx 635$ Jto 3 significant figures

Conclusion

The body gains as kinetic energy exactly the work of the net force on it: what the push puts in less what friction takes out.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_pushed_body_gains_kinetic_energy do4  kind :numerical5  describes "A body of mass %{mass} initially at rest moves under the action of an applied horizontal force of " \6            "%{force} on a table with coefficient of kinetic friction = 0.1. Taking g as %{gravity}, compute the " \7            "%{gain} in %{time}, in joules."8  given mass: "2 kg"9  given force: "7 N"10  given gravity: "9.8 m/s2"11  given time: "10 s"12  asks gain: "change in kinetic energy of the body"13end

solution.rb

1require_relative "../questions/a_pushed_body_gains_kinetic_energy"2require_relative "../science/work_energy_and_power/kinetic_energy"34solve :a_pushed_body_gains_kinetic_energy, using: KineticEnergy do5  objects :body, :hand, :table6  axes :x7  body.goes(:sliding)89  take mass: body.mass10  take force: body.force(by: hand).magnitude11  take gravity: g12  take time: sliding.time_taken1314  assume body.initial_position => "0 m"15  deduce_from "initially at rest", body.initial_velocity => "0 m/s"16  deduce_from "applied horizontal force", body.force(by: hand).direction => x17  deduce_from "coefficient of kinetic friction = 0.1", body.force(by: table) => -0.1 * body.mass * g1819  gain = body.change_of_kinetic_energy20  answer gain:, in: "J", figures: 32122  conclusion "The body gains as kinetic energy exactly the work of the net force on it: what the push puts in " \23             "less what friction takes out."24end

playground.rb

1require_relative "../solutions/a_pushed_body_gains_kinetic_energy"23play_with :a_pushed_body_gains_kinetic_energy do4  title "A Pushed Body Gains Kinetic Energy"5end

The laws it drew on

quantity	written	units	within	from
volume	$volume$	—	—	Vectors
mass	$m$	kg	0 … ∞	Vectors
speed	$speed$	—	—	Vectors
density	$density$	—	—	Vectors
number of moles	$n$	—	—	Vectors
angular frequency	${angular}_{frequency}$	—	—	Vectors
work	$W_{F}$	J	—	Vectors
current	$current$	—	—	Vectors
temperature	$temperature$	—	—	Vectors
pressure	$pressure$	—	—	Vectors
time	$t$	s	0 … 1e+05	Vectors
power	$power$	—	—	Vectors
total path length	$s$	—	—	Vectors
energy	$energy$	—	—	Vectors
gravitational potential	${gravitational}_{potential}$	—	—	Vectors
coefficient of friction	$μ$	—	—	Vectors
charge	$charge$	—	—	Vectors
acceleration	$a$	m/s²	—	Vectors
velocity	$v$	m/s	—	Vectors
displacement	$Δ x$	m	—	Vectors
angular velocity	${angular}_{velocity}$	—	—	Vectors
force	$F$	N	—	Vectors
angular momentum	${angular}_{momentum}$	—	—	Vectors
linear momentum	$p_{lin}$	—	—	Vectors
electric field	${electric}_{field}$	—	—	Vectors
average velocity	$v_{av}$	m/s	—	Vectors
magnetic moment	${magnetic}_{moment}$	—	—	Vectors
relative velocity	$v_{AB}$	—	—	Vectors
impulse	$impulse$	—	—	Vectors
first vector	$A$	—	0 … ∞	Vectors
second vector	$B$	—	0 … ∞	Vectors
angle between	$φ$	—	0 … 3.142	Vectors
resultant	$R$	—	0 … ∞	Vectors
resultant direction	$β$	—	0 … 3.142	Vectors
direction	$α$	—	-3.142 … 3.142	Vectors
x component	$A_{x}$	—	—	Vectors
y component	$A_{y}$	—	—	Vectors
position	$x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
change of velocity	$Δ v$	m/s	—	Acceleration
average acceleration	$a_{av}$	m/s²	—	Acceleration
starting velocity	$v_{0}$	m/s	—	Acceleration
final velocity	$v$	m/s	—	Acceleration
gravity	$g$	m/s²	—	UniformAcceleration
net force	$F_{net}$	N	—	NewtonsSecondLaw
weight	$W$	N	—	NewtonsSecondLaw
work by gravity	$W_{g}$	J	—	Work
net work	$W_{net}$	J	—	Work
kinetic energy	$K$	J	—	KineticEnergy
change of kinetic energy	$Δ K$	J	—	KineticEnergy
displacement of body during sliding along x	$Δ x$	m	—
position of body at end of sliding along x	$x_{1}$	m	—
position of body at start of sliding along x	$x_{0}$	m	—
average velocity of body during sliding along x	$v_{av}$	m/s	—
time taken of sliding	$Δ t$	s	0 … 1e+05
average speed of body during sliding	$c_{av}$	m/s	—
path length of body during sliding	$d$	m	—
change of velocity of body during sliding along x	$Δ v$	m/s	—
velocity of body at end of sliding from ground along x	$v_{1}$	m/s	—
velocity of body at start of sliding from ground along x	$v_{0}$	m/s	—
average acceleration of body during sliding along x	$a_{av}$	m/s²	—
acceleration of body during sliding along x	$a$	m/s²	—
net force on body during sliding along x	$F_{net}$	N	—
mass of body	$m$	kg	0 … ∞
mass of hand	$m_{h}$	kg	0 … ∞
mass of table	$m_{t}$	kg	0 … ∞
weight on ground	$\| W_{g} \|$	N	—
weight on body	$\| W \|$	N	—
weight on hand	$\| W_{h} \|$	N	—
weight on table	$\| W_{t} \|$	N	—
force on body by hand during sliding	$\| F_{h} \|$	N	—
force on body by table during sliding	$\| F_{t} \|$	N	—
force on body by hand during sliding along x	$F_{h}$	N	—
force on body by table during sliding along x	$F_{t}$	N	—
work on body by hand during sliding	$W_{F,h}$	J	—
work on body by table during sliding	$W_{F,t}$	J	—
net work on body during sliding	$W_{net}$	J	—
kinetic energy of body at start of sliding	$K_{0}$	J	—
kinetic energy of body at end of sliding	$K_{1}$	J	—
change of kinetic energy of body during sliding	$Δ K$	J	—
displacement of body during sliding	$\| Δ x \|$	m	—
position of body at end of sliding	$\| x_{1} \|$	m	—
position of body at start of sliding	$\| x_{0} \|$	m	—
average velocity of body during sliding	$\| v_{av} \|$	m/s	—
change of velocity of body during sliding	$\| Δ v \|$	m/s	—
velocity of body at end of sliding from ground	$\| v_{1} \|$	m/s	—
velocity of body at start of sliding from ground	$\| v_{0} \|$	m/s	—
average acceleration of body during sliding	$\| a_{av} \|$	m/s²	—
acceleration of body during sliding	$\| a \|$	m/s²	—
net force on body during sliding	$\| F_{net} \|$	N	—

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

v_{0} \equiv v_{i} (start of motion)

starting velocity

v \equiv v_{i} (end of motion)

final velocity

t \equiv Δ t_{m}

time

A = \sqrt{{A_{x}}^{2} + {A_{y}}^{2}}

magnitude from components

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

a_{i} (part 1) = a_{i} (whole)

steady through its parts

m_{b} \neq 0

has mass?

R^{2} = A^{2} + B^{2} + 2 \cdot A \cdot B \cdot \cos φ

law of cosines

R \cdot \sin β = B \cdot \sin φ

law of sines

R \cdot \cos β = A + B \cdot \cos φ

along the first

A_{x} = A \cdot \cos α

resolved along x

A_{y} = A \cdot \sin α

resolved along y

Δ x = x_{1} - x_{0}

change of position(sliding, x)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(sliding, x)

c_{av} = \frac{d}{Δ t}

path a time(sliding, x)

Δ v = v_{1} - v_{0}

velocity gained(sliding, x)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(sliding, x)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(sliding, x)

x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}

position after a time(sliding, x)

{v_{1}}^{2} = {v_{0}}^{2} + 2 \cdot a \cdot (x_{1} - x_{0})

velocity after a distance(sliding, x)

x_{1} - x_{0} = \frac{(v_{1} + v_{0}) \cdot Δ t}{2}

mean of the velocities(sliding, x)

F_{net} = m \cdot a

second law(sliding, x)

F_{net} = F_{h} + F_{t} + 0

net force is the resultant(body, sliding, x)

W_{F,h} = F_{h} \cdot Δ x

work of a steady force(body, sliding, 1)

W_{F,t} = F_{t} \cdot Δ x

work of a steady force(body, sliding, 2)

W_{net} = F_{net} \cdot Δ x

work of the net force(body, sliding)

K_{0} = \frac{m \cdot {v_{0}}^{2}}{2}

kinetic energy of a body(body, sliding, 1)

K_{1} = \frac{m \cdot {v_{1}}^{2}}{2}

kinetic energy of a body(body, sliding, 2)

Δ K = K_{1} - K_{0}

kinetic energy gained(body, sliding)

K_{1} - K_{0} = W_{F,h} + W_{F,t}

work energy theorem(body, sliding)

| Δ x | = \sqrt{{Δ x}^{2}}

magnitude from components(displacement of body during sliding)

| x_{1} | = \sqrt{{x_{1}}^{2}}

magnitude from components(position of body at end of sliding)

| x_{0} | = \sqrt{{x_{0}}^{2}}

magnitude from components(position of body at start of sliding)

| v_{av} | = \sqrt{{v_{av}}^{2}}

magnitude from components(average velocity of body during sliding)

| Δ v | = \sqrt{{Δ v}^{2}}

magnitude from components(change of velocity of body during sliding)

| v_{1} | = \sqrt{{v_{1}}^{2}}

magnitude from components(velocity of body at end of sliding from ground)

| v_{0} | = \sqrt{{v_{0}}^{2}}

magnitude from components(velocity of body at start of sliding from ground)

| a_{av} | = \sqrt{{a_{av}}^{2}}

magnitude from components(average acceleration of body during sliding)

| a | = \sqrt{a^{2}}

magnitude from components(acceleration of body during sliding)

| F_{net} | = \sqrt{{F_{net}}^{2}}

magnitude from components(net force on body during sliding)

| F_{h} | = \sqrt{{F_{h}}^{2}}

magnitude from components(force on body by hand during sliding)

| F_{t} | = \sqrt{{F_{t}}^{2}}

magnitude from components(force on body by table during sliding)

F_{h} = | F_{h} |

applied horizontal force

F_{t} = −0.1 \cdot m \cdot g

coefficient of kinetic friction = 0.1

m \neq 0

has mass?(body)

m_{h} \neq 0

has mass?(hand)

m_{t} \neq 0

has mass?(table)

F_{net} = 0

free of forces?(body, sliding)