play phryzby with physics

Stopped by a Steady Force

solved NewtonsSecondLaw

Q. A body of 20 kgm=20kg is moving at 15 m/sv0=15m/s when a steady force of 50 N|Fnet|=50N starts to act against its motionFnet=−|Fnet|. How long does it take to stopv1=0m/s, in seconds?

bodym = 20 kgv₀ = 15 m/sv = 0 m/sΔt = ?aF = 50 Nv
Inputs
m
20 kg
v0
15.0 m/s
|Fnet|
50 N
Outputs
timeΔt?

How it's solved

Solved with NewtonsSecondLaw

Given

  • m=20kg
  • v0=15m/s
  • |Fnet|=50N
  • v1=0m/s from “to stop”
  • Fnet=−|Fnet| from “against its motion”

Find Δt.

Working
  1. against its motionFnet=−|Fnet|
  2. Fnet=−50
  3. Fnet=−50N
  4. second law(stopping, x)Fnet=m⋅a
  5. (−50)=20⋅a
  6. a=−2.5m/s²
  7. velocity after a time(stopping, x)v1=v0+a⋅Δt
  8. 0=15+(−2.5)⋅Δt
  9. Δt=6.00 s
Conclusion

The force slows it at a steady rate, the force over the mass, so the time to stop is the speed it has to lose over that rate.

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 :stopped_by_a_steady_force do4  kind :numerical5  describes "A body of %{mass} is moving at %{speed} when a steady force of %{force} starts to act against " \6            "its motion. %{time}, in seconds?"7  given mass: "20 kg"8  given speed: "15 m/s"9  given force: "50 N"10  asks time: "How long does it take to stop"11end
solution.rb
1require_relative "../questions/stopped_by_a_steady_force"2require_relative "../science/laws_of_motion/newtons_second_law"34solve :stopped_by_a_steady_force, using: NewtonsSecondLaw do5  objects :body6  axes :x7  body.goes(:stopping)89  take mass: body.mass10  take speed: body.initial_velocity11  take force: body.net_force.magnitude1213  deduce_from "against its motion", body.net_force.direction => -x14  deduce_from "to stop", body.final_velocity => "0 m/s"1516  time = stopping.time_taken17  answer time:, in: "s"1819  conclusion "The force slows it at a steady rate, the force over the mass, so the time to stop is " \20             "the speed it has to lose over that rate."21end
playground.rb
1require_relative "../solutions/stopped_by_a_steady_force"23play_with :stopped_by_a_steady_force do4  title "Stopped by a Steady Force"5end
The laws it drew on
quantitywrittenunitswithinfrom
volumevolume——Vectors
massmkg0 … ∞Vectors
speedspeed——Vectors
densitydensity——Vectors
number of molesn——Vectors
angular frequencyangularfrequency——Vectors
workwork——Vectors
currentcurrent——Vectors
temperaturetemperature——Vectors
pressurepressure——Vectors
timets0 … 1e+05Vectors
powerpower——Vectors
total path lengths——Vectors
energyenergy——Vectors
gravitational potentialgravitationalpotential——Vectors
coefficient of frictionμ——Vectors
chargecharge——Vectors
accelerationam/s²—Vectors
velocityvm/s—Vectors
displacementΔxm—Vectors
angular velocityangularvelocity——Vectors
forceFN—Vectors
angular momentumangularmomentum——Vectors
linear momentumplin——Vectors
electric fieldelectricfield——Vectors
average velocityvavm/s—Vectors
magnetic momentmagneticmoment——Vectors
relative velocityvAB——Vectors
impulseimpulse——Vectors
first vectorA—0 … ∞Vectors
second vectorB—0 … ∞Vectors
angle betweenφ—0 … 3.142Vectors
resultantR—0 … ∞Vectors
resultant directionβ—0 … 3.142Vectors
directionα—-3.142 … 3.142Vectors
x componentAx——Vectors
y componentAy——Vectors
positionxm—Velocity
path lengthdm—Velocity
time takenΔts0 … 1e+05Velocity
average speedcavm/s—Velocity
starting positionx0m—Velocity
final positionxm—Velocity
change of velocityΔvm/s—Acceleration
average accelerationaavm/s²—Acceleration
starting velocityv0m/s—Acceleration
final velocityvm/s—Acceleration
gravitygm/s²—UniformAcceleration
net forceFnetN—NewtonsSecondLaw
weightWN—NewtonsSecondLaw
displacement of body during stopping along xΔxm—
position of body at end of stopping along xx1m—
position of body at start of stopping along xx0m—
average velocity of body during stopping along xvavm/s—
time taken of stoppingΔts0 … 1e+05
average speed of body during stoppingcavm/s—
path length of body during stoppingdm—
change of velocity of body during stopping along xΔvm/s—
velocity of body at end of stopping from ground along xv1m/s—
velocity of body at start of stopping from ground along xv0m/s—
average acceleration of body during stopping along xaavm/s²—
acceleration of body during stopping along xam/s²—
net force on body during stopping along xFnetN—
mass of bodymkg0 … ∞
weight on ground|Wg|N—
weight on body|W|N—
displacement of body during stopping|Δx|m—
position of body at end of stopping|x1|m—
position of body at start of stopping|x0|m—
average velocity of body during stopping|vav|m/s—
change of velocity of body during stopping|Δv|m/s—
velocity of body at end of stopping from ground|v1|m/s—
velocity of body at start of stopping from ground|v0|m/s—
average acceleration of body during stopping|aav|m/s²—
acceleration of body during stopping|a|m/s²—
net force on body during stopping|Fnet|N—
x0≡xi(start of motion)
starting position
x≡xi(end of motion)
final position
v0≡vi(start of motion)
starting velocity
v≡vi(end of motion)
final velocity
t≡Δtm
time
A=Ax2+Ay2
magnitude from components
di(whole)=di(part 1)+di(part 2)
path in parts
Δtw=Δtp+Δtp
time in parts
ai(part 1)=ai(whole)
steady through its parts
mb≠0
has mass?
R2=A2+B2+2⋅A⋅B⋅cos⁡φ
law of cosines
R⋅sin⁡β=B⋅sin⁡φ
law of sines
R⋅cos⁡β=A+B⋅cos⁡φ
along the first
Ax=A⋅cos⁡α
resolved along x
Ay=A⋅sin⁡α
resolved along y
Δx=x1-x0
change of position(stopping, x)
vav=ΔxΔt
displacement a time(stopping, x)
cav=dΔt
path a time(stopping, x)
Δv=v1-v0
velocity gained(stopping, x)
aav=ΔvΔt
velocity gained a time(stopping, x)
v1=v0+a⋅Δt
velocity after a time(stopping, x)
x1=x0+v0⋅Δt+a⋅Δt22
position after a time(stopping, x)
v12=v02+2⋅a⋅(x1-x0)
velocity after a distance(stopping, x)
x1-x0=(v1+v0)⋅Δt2
mean of the velocities(stopping, x)
Fnet=m⋅a
second law(stopping, x)
|Δx|=Δx2
magnitude from components(displacement of body during stopping)
|x1|=x12
magnitude from components(position of body at end of stopping)
|x0|=x02
magnitude from components(position of body at start of stopping)
|vav|=vav2
magnitude from components(average velocity of body during stopping)
|Δv|=Δv2
magnitude from components(change of velocity of body during stopping)
|v1|=v12
magnitude from components(velocity of body at end of stopping from ground)
|v0|=v02
magnitude from components(velocity of body at start of stopping from ground)
|aav|=aav2
magnitude from components(average acceleration of body during stopping)
|a|=a2
magnitude from components(acceleration of body during stopping)
|Fnet|=Fnet2
magnitude from components(net force on body during stopping)
Fnet=−|Fnet|
against its motion
m≠0
has mass?(body)
Fnet=0
free of forces?(body, stopping)