Pushed Back for Twenty-Five Seconds

solved NewtonsSecondLaw

Q. A body of mass 0.40 kg $m = 0.40$ kg moving initially with a constant speed of 10 m/s $v (applied) = 10$ m/s to the north is subject to a constant force of 8.0 N $| F_{p} | = 8.0$ N directed towards the south $F_{p} = - | F_{p} |$ for 30 s. Take the instant the force is applied to be t = 0, and the position of the body at that time to be x = 0 $x (applied) = 0$ m. What is its position at t = 25 s $Δ t = 25$ s, in kilometres?

Inputs

mass

m

0.4 kg

speed

v

10.0 m/s

force

| F_{p} |

8 N

time

Δ t

25.00 s

Outputs

position

x

How it's solved

Solved with NewtonsSecondLaw

Given

$m = 0.40$ kg
$v (applied) = 10$ m/s
$| F_{p} | = 8.0$ N
$Δ t = 25$ s
$x (applied) = 0$ m from “x = 0”
$F_{p} = - | F_{p} |$ from “directed towards the south”

Find $x (later)$ .

Working

directed towards the south $F_{p} = - | F_{p} |$
$F_{p} = - 8$
$F_{p} = −8$ N
net force is the resultant(body, so far, x) $F_{net} = F_{p} + 0$
$F_{net} = (−8) + 0$
$F_{net} = −8$ N
second law(so far, x) $F_{net} = m \cdot a$
$(−8) = 0.4 \cdot a$
$a = −20$ m/s²
position after a time(so far, x) $x (later) = x (applied) + v (applied) \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$x (later) = 0 + 10 \cdot 25 + \frac{(−20) \cdot 25^{2}}{2}$
$x (later) = −6$ km
$x (later) = −6000$ m $= −6$ km
$x (later) \approx −6$ kmto 1 significant figure

Conclusion

The force pulls the body back the whole time: it stops, turns round and is soon far to the south of where the force first caught it.

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 :pushed_back_for_twenty_five_seconds do4  kind :numerical5  describes "A body of mass %{mass} moving initially with a constant speed of %{speed} to the north is subject to a " \6            "constant force of %{force} directed towards the south for 30 s. Take the instant the force is " \7            "applied to be t = 0, and the position of the body at that time to be x = 0. %{position} at t = %{time}, " \8            "in kilometres?"9  given mass: "0.40 kg"10  given speed: "10 m/s"11  given force: "8.0 N"12  given time: "25 s"13  asks position: "What is its position"14end

solution.rb

1require_relative "../questions/pushed_back_for_twenty_five_seconds"2require_relative "../science/laws_of_motion/newtons_second_law"34solve :pushed_back_for_twenty_five_seconds, using: NewtonsSecondLaw do5  objects :body, :pusher6  axes :x7  instants :applied, :later8  body.goes(:so_far, during: applied..later)910  take mass: body.mass11  take speed: body.velocity(at: applied)12  take force: body.force(by: pusher).magnitude13  take time: so_far.time_taken1415  deduce_from "directed towards the south", body.force(by: pusher).direction => -x16  deduce_from "x = 0", body.position(at: applied) => "0 m"1718  position = body.position(at: later)19  answer position:, in: "km", figures: 12021  conclusion "The force pulls the body back the whole time: it stops, turns round and is soon far to the " \22             "south of where the force first caught it."23end

playground.rb

1require_relative "../solutions/pushed_back_for_twenty_five_seconds"23play_with :pushed_back_for_twenty_five_seconds do4  title "Pushed Back for Twenty-Five Seconds"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	$work$	—	—	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
displacement of body during so far along x	$Δ x$	m	—
position of body at later along x	$x (later)$	m	—
position of body at applied along x	$x (applied)$	m	—
average velocity of body during so far along x	$v_{av}$	m/s	—
time taken of so far	$Δ t$	s	0 … 1e+05
average speed of body during so far	$c_{av}$	m/s	—
path length of body during so far	$d$	m	—
change of velocity of body during so far along x	$Δ v$	m/s	—
velocity of body at later from ground along x	$v (later)$	m/s	—
velocity of body at applied from ground along x	$v (applied)$	m/s	—
average acceleration of body during so far along x	$a_{av}$	m/s²	—
acceleration of body during so far along x	$a$	m/s²	—
net force on body during so far along x	$F_{net}$	N	—
mass of body	$m$	kg	0 … ∞
mass of pusher	$m_{p}$	kg	0 … ∞
weight on ground	$\| W_{g} \|$	N	—
weight on body	$\| W \|$	N	—
weight on pusher	$\| W_{p} \|$	N	—
force on body by pusher during so far	$\| F_{p} \|$	N	—
force on body by pusher during so far along x	$F_{p}$	N	—
displacement of body during so far	$\| Δ x \|$	m	—
position of body at later	$\| x (later) \|$	m	—
position of body at applied	$\| x (applied) \|$	m	—
average velocity of body during so far	$\| v_{av} \|$	m/s	—
change of velocity of body during so far	$\| Δ v \|$	m/s	—
velocity of body at later from ground	$\| v (later) \|$	m/s	—
velocity of body at applied from ground	$\| v (applied) \|$	m/s	—
average acceleration of body during so far	$\| a_{av} \|$	m/s²	—
acceleration of body during so far	$\| a \|$	m/s²	—
net force on body during so far	$\| F_{net} \|$	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 (later) - x (applied)

change of position(so far, x)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(so far, x)

c_{av} = \frac{d}{Δ t}

path a time(so far, x)

Δ v = v (later) - v (applied)

velocity gained(so far, x)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(so far, x)

v (later) = v (applied) + a \cdot Δ t

velocity after a time(so far, x)

x (later) = x (applied) + v (applied) \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}

position after a time(so far, x)

{v (later)}^{2} = {v (applied)}^{2} + 2 \cdot a \cdot (x (later) - x (applied))

velocity after a distance(so far, x)

x (later) - x (applied) = \frac{(v (later) + v (applied)) \cdot Δ t}{2}

mean of the velocities(so far, x)

F_{net} = m \cdot a

second law(so far, x)

F_{net} = F_{p} + 0

net force is the resultant(body, so far, x)

| Δ x | = \sqrt{{Δ x}^{2}}

magnitude from components(displacement of body during so far)

| x (later) | = \sqrt{{x (later)}^{2}}

magnitude from components(position of body at later)

| x (applied) | = \sqrt{{x (applied)}^{2}}

magnitude from components(position of body at applied)

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

magnitude from components(average velocity of body during so far)

| Δ v | = \sqrt{{Δ v}^{2}}

magnitude from components(change of velocity of body during so far)

| v (later) | = \sqrt{{v (later)}^{2}}

magnitude from components(velocity of body at later from ground)

| v (applied) | = \sqrt{{v (applied)}^{2}}

magnitude from components(velocity of body at applied from ground)

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

magnitude from components(average acceleration of body during so far)

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

magnitude from components(acceleration of body during so far)

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

magnitude from components(net force on body during so far)

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

magnitude from components(force on body by pusher during so far)

F_{p} = - | F_{p} |

directed towards the south

m \neq 0

has mass?(body)

m_{p} \neq 0

has mass?(pusher)

F_{net} = 0

free of forces?(body, so far)