A Steady Push Forward

Q. A steady force on a body of 3.0 kg takes it from 2.0 m/s to 3.5 m/s in 25 s, without turning it. How strong is the force, in newtons? expects 0.18 N

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 steady force on a body of 3.0 kg $m = 3.0$ kg takes it from 2.0 m/s $v_{0} = 2.0$ m/s to 3.5 m/s $v_{1} = 3.5$ m/s in 25 s $Δ t = 25$ s, without turning it. How strong is the force, in newtons?

The playground

Inputs

mass

m

3 kg

initial speed

v_{0}

2.0 m/s

final speed

v_{1}

3.5 m/s

time

Δ t

25.00 s

Outputs

force

F_{net}

The solution

Solved with NewtonsSecondLaw

Given

$m = 3.0$ kg
$v_{0} = 2.0$ m/s
$v_{1} = 3.5$ m/s
$Δ t = 25$ s

Find $F_{net}$ .

Working

velocity after a time(speeding up) $v_{1} = v_{0} + a \cdot Δ t$
$3.5 = 2 + a \cdot 25$
$a = 0.06$ m/s²
second law(speeding up) $F_{net} = m \cdot a$
$F_{net} = 3 \cdot 0.06$
$F_{net} = 0.18$ N
$F_{net} \approx 0.18$ Nto 2 significant figures

Conclusion

The force is the mass times the speed it gains each second, and it points the way the body is already going.

The Ruby

1require_relative "../physics"23question :a_steady_push_forward do4  kind :numerical5  describes "A steady force on a body of %{mass} takes it from %{initial_speed} to %{final_speed} in %{time}, without " \6            "turning it. %{force}, in newtons?"7  given mass: "3.0 kg"8  given initial_speed: "2.0 m/s"9  given final_speed: "3.5 m/s"10  given time: "25 s"11  asks force: "How strong is the force"12end

1require_relative "../questions/a_steady_push_forward"2require_relative "../science/laws_of_motion/newtons_second_law"34solve :a_steady_push_forward, using: NewtonsSecondLaw do5  objects :body6  body.goes(:speeding_up)78  take mass: body.mass9  take initial_speed: body.initial_velocity10  take final_speed: body.final_velocity11  take time: speeding_up.time_taken1213  force = body.net_force14  answer force:, in: "N", figures: 21516  conclusion "The force is the mass times the speed it gains each second, and it points the way " \17             "the body is already going."18end

1require_relative "../solutions/a_steady_push_forward"23play_with :a_steady_push_forward do4  title "A Steady Push Forward"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 speeding up	$Δ x$	m	—
position of body at end of speeding up	$x_{1}$	m	—
position of body at start of speeding up	$x_{0}$	m	—
average velocity of body during speeding up	$v_{av}$	m/s	—
time taken of speeding up	$Δ t$	s	0 … 1e+05
average speed of body during speeding up	$c_{av}$	m/s	—
path length of body during speeding up	$d$	m	—
change of velocity of body during speeding up	$Δ v$	m/s	—
velocity of body at end of speeding up from ground	$v_{1}$	m/s	—
velocity of body at start of speeding up from ground	$v_{0}$	m/s	—
average acceleration of body during speeding up	$a_{av}$	m/s²	—
acceleration of body during speeding up	$a$	m/s²	—
net force on body during speeding up	$F_{net}$	N	—
mass of body	$m$	kg	0 … ∞
weight on ground	$W_{g}$	N	—
weight on body	$W$	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(speeding up)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(speeding up)

c_{av} = \frac{d}{Δ t}

path a time(speeding up)

Δ v = v_{1} - v_{0}

velocity gained(speeding up)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(speeding up)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(speeding up)

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

position after a time(speeding up)

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

velocity after a distance(speeding up)

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

mean of the velocities(speeding up)

F_{net} = m \cdot a

second law(speeding up)

m \neq 0

has mass?(body)

F_{net} = 0

free of forces?(body, speeding up)