The Work of a Steady Push

Q. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Taking g as 9.8 m/s², compute the work done by the applied force in 10 s, in joules. expects 882 J

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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 work done by the applied force in 10 s $Δ t = 10$ s, in joules.

The playground

Inputs

mass

m

2 kg

force

| F_{h} |

7 N

gravity

g

9.8 m/s²

time

Δ t

10.00 s

Outputs

work

W_{F,h}

The solution

Solved with Work

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 $W_{F,h}$ .

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
position after a time(sliding, x) $x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$x_{1} = 0 + 0 \cdot 10 + \frac{2.52 \cdot 10^{2}}{2}$
$x_{1} = 126$ m
change of position(sliding, x) $Δ x = x_{1} - x_{0}$
$Δ x = 126 - 0$
$Δ x = 126$ m
work of a steady force(body, sliding, 1) $W_{F,h} = F_{h} \cdot Δ x$
$W_{F,h} = 7 \cdot 126$
$W_{F,h} = 882$ J
$W_{F,h} \approx 882$ Jto 3 significant figures

Conclusion

The push stays the same along the whole path, so its work is the push times the distance the body covers.

The Ruby

1require_relative "../physics"23question :the_work_of_a_steady_push 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            "%{work} 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 work: "work done by the applied force"13end

1require_relative "../questions/the_work_of_a_steady_push"2require_relative "../science/work_energy_and_power/work"34solve :the_work_of_a_steady_push, using: Work 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  work = body.work(by: hand)20  answer work:, in: "J", figures: 32122  conclusion "The push stays the same along the whole path, so its work is the push times the distance the body covers."23end

1require_relative "../solutions/the_work_of_a_steady_push"23play_with :the_work_of_a_steady_push do4  title "The Work of a Steady Push"5end

The laws

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
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	—
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)

| 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)