play phryzby with physics

A Pebble Thrown Up

solved NewtonsSecondLaw

Q. A pebble of mass 0.05 kgm=0.05kg is thrown vertically upwards. Ignore air resistanceFa=0N, and take g asW=−|W| 10 m/s²g=10m/s². What is the magnitude of the net force on the pebble during its upward motion, in newtons?

pebbleaweight
Inputs
m
0.05 kg
g
10 m/s²
Outputs
force|Fnet|?

How it's solved

Solved with NewtonsSecondLaw

Given

  • m=0.05kg
  • g=10m/s²
  • Fa=0N from “Ignore air resistance”
  • W=−|W| from “take g as”

Find |Fnet|.

Working
  1. weight of a body(pebble)|W|=m⋅g
  2. |W|=0.05⋅10
  3. |W|=0.5N
  4. take g asW=−|W|
  5. W=−0.5
  6. W=−0.5N
  7. net force is the resultant(pebble, rising, y)Fnet=Fa+W
  8. Fnet=0+(−0.5)
  9. Fnet=−0.5N
  10. magnitude from components(net force on pebble during rising)|Fnet|=Fnet2
  11. |Fnet|=(−0.5)2
  12. |Fnet|=0.5N
Conclusion

With the air left out, the only force on the pebble is its weight, downward, whether it is going up, coming down or still for an instant at the top.

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 :a_pebble_thrown_up do4  kind :numerical5  describes "A pebble of mass %{mass} is thrown vertically upwards. Ignore air resistance, and take g as %{gravity}. " \6            "%{force} during its upward motion, in newtons?"7  given mass: "0.05 kg"8  given gravity: "10 m/s2"9  asks force: "What is the magnitude of the net force on the pebble"10end
solution.rb
1require_relative "../questions/a_pebble_thrown_up"2require_relative "../science/laws_of_motion/newtons_second_law"34solve :a_pebble_thrown_up, using: NewtonsSecondLaw do5  objects :pebble, :air6  axes :y7  pebble.goes(:rising)8  deduce_from "take g as", pebble.weight.direction => -y910  take mass: pebble.mass11  take gravity: g1213  deduce_from "Ignore air resistance", pebble.force(by: air) => "0 N"1415  force = pebble.net_force.magnitude16  answer force:, in: "N"1718  conclusion "With the air left out, the only force on the pebble is its weight, downward, whether it is " \19             "going up, coming down or still for an instant at the top."20end
playground.rb
1require_relative "../solutions/a_pebble_thrown_up"23play_with :a_pebble_thrown_up do4  title "A Pebble Thrown Up"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 pebble during rising along yΔxm—
position of pebble at end of rising along yx1m—
position of pebble at start of rising along yx0m—
average velocity of pebble during rising along yvavm/s—
time taken of risingΔts0 … 1e+05
average speed of pebble during risingcavm/s—
path length of pebble during risingdm—
change of velocity of pebble during rising along yΔvm/s—
velocity of pebble at end of rising from ground along yv1m/s—
velocity of pebble at start of rising from ground along yv0m/s—
average acceleration of pebble during rising along yaavm/s²—
acceleration of pebble during rising along yam/s²—
net force on pebble during rising along yFnetN—
mass of pebblemkg0 … ∞
mass of airmakg0 … ∞
weight on ground|Wg|N—
weight on pebble|W|N—
weight on air|Wa|N—
force on pebble by air during rising|Fa|N—
weight on pebble along yWN—
force on pebble by air during rising along yFaN—
displacement of pebble during rising|Δx|m—
position of pebble at end of rising|x1|m—
position of pebble at start of rising|x0|m—
average velocity of pebble during rising|vav|m/s—
change of velocity of pebble during rising|Δv|m/s—
velocity of pebble at end of rising from ground|v1|m/s—
velocity of pebble at start of rising from ground|v0|m/s—
average acceleration of pebble during rising|aav|m/s²—
acceleration of pebble during rising|a|m/s²—
net force on pebble during rising|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(rising, y)
vav=ΔxΔt
displacement a time(rising, y)
cav=dΔt
path a time(rising, y)
Δv=v1-v0
velocity gained(rising, y)
aav=ΔvΔt
velocity gained a time(rising, y)
v1=v0+a⋅Δt
velocity after a time(rising, y)
x1=x0+v0⋅Δt+a⋅Δt22
position after a time(rising, y)
v12=v02+2⋅a⋅(x1-x0)
velocity after a distance(rising, y)
x1-x0=(v1+v0)⋅Δt2
mean of the velocities(rising, y)
Fnet=m⋅a
second law(rising, y)
|W|=m⋅g
weight of a body(pebble)
Fnet=Fa+W
net force is the resultant(pebble, rising, y)
|Δx|=Δx2
magnitude from components(displacement of pebble during rising)
|x1|=x12
magnitude from components(position of pebble at end of rising)
|x0|=x02
magnitude from components(position of pebble at start of rising)
|vav|=vav2
magnitude from components(average velocity of pebble during rising)
|Δv|=Δv2
magnitude from components(change of velocity of pebble during rising)
|v1|=v12
magnitude from components(velocity of pebble at end of rising from ground)
|v0|=v02
magnitude from components(velocity of pebble at start of rising from ground)
|aav|=aav2
magnitude from components(average acceleration of pebble during rising)
|a|=a2
magnitude from components(acceleration of pebble during rising)
|Fnet|=Fnet2
magnitude from components(net force on pebble during rising)
|W|=W2
magnitude from components(weight on pebble)
|Fa|=Fa2
magnitude from components(force on pebble by air during rising)
W=−|W|
take g as
m≠0
has mass?(pebble)
ma≠0
has mass?(air)
Fnet=0
free of forces?(pebble, rising)