Pushed Back for Twenty-Five Seconds

Q. A body of mass 0.40 kg moving initially with a constant speed of 10 m/s to the north is subject to a constant force of 8.0 N directed towards the south 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. What is its position at t = 25 s, in kilometres? expects -6 km

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

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

The solution

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 Ruby

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

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

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

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)