A 1-kg object 2 m above the floor has a gravitational potential energy of 3 J

A 1-kg object 2 m above the floor in a regular building at the Earth has a gravitational potential energy of 3 J.
How much gravitational potential energy does the object have when it is 4 m above the floor?

Given Data:
m = 1kg
y₁ = 2m
U₁ = 3J
y₂ = 4m
U₂ = ?

Useful Physics Formulas

U = mgh
ΔU = Δ(mgh) = mgΔh

Solution
ΔU = Δ(mgh) = mgΔh = mg(y₂ - y₁)
ΔU = U₂ - U₁
U₂ = U₁ + mg(y₂ - y₁)

U₂ = 3J + 1kg · 10ᵐ/s² · (4m - 2m) = 3J + 20J = 23J

Problem's answer is 23 J.

Δ·⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ⁱⁿᴬᴭᴮᴯᴱᴲᴳᴴᴵᴶᴷᴸᴹᴺᴻᴼᴽᴾᴿᵀᵁᵂᵃᵄᵅᵆᵇᵉᵊᵋᵌᵍᵏᵐᵑᵒᵓᵖᵗᵘᵚᵛᵝᵞᵟᵠᵡ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛᵢᵣᵤᵥᵦᵧᵨᵩᵪ½↉⅓⅔¼¾⅕⅖⅗⅘⅙⅚⅐⅛⅜⅝⅞⅑⅒⅟℀℁℅℆

BMCC Students, Please Repeat Your Submission for the Quiz 2

BMCC Students, Please repeat your submission for the quiz 2! Unfortunately for technical reasons your previous submissions of the quiz 2 were not recorded.

Physics Problem - A spring is fixed vertically to the floor

A #spring is fixed #vertically to the floor.
A 10-kg #box is then #pressed down such that the spring is #compressed https://t.co/Z5fLcbq2LX

— Physics I (@PHY210) October 5, 2016

A spring is fixed vertically to the floor.
A 10.0-kg box is then pressed down such that the spring is compressed 0.500 m beyond its equilibrium length.
The box is then released and it travels vertically upward to a maximum height of 2.40 m above its release point.
Assuming the work done against drag (air resistance) is 120. J, what is the spring constant of the spring?
Given Data:
M= 10.0-kg
x= 0.5 m
h=2.40 m
Wᵣ =120 J
k=?
Solution:
½kx²=mgh+Wᵣ Law of Energy Conservation
k=2(mgh+Wᵣ )/x² Symbolical Solution
2*(10kg*9.82m/s^2*2.40m+120J)/(0.5m)^2 in N/m Code of Google Calculator
2845.44 N / m Google Calculator's result
k=2840 N/m

How much does the spring compress?

The force on this mass when it is located at r =... m is

The potential energy is given by the function U(x) = kx⁻¹, where k is a constant, and variable x is the position, the coordinate. (a) Derive the function of the coordinate x, which describes the force produced by this potential energy.

F(x) = - d U(x) / dx

F(x) = - d kx⁻¹ / dx = kx⁻²

F(x) = kx⁻²

(b) Let this potential energy is obtained by the body with the mass of 10kg. The constant is known: k = -3.98 × 10¹⁵ Jm.
Find the force on this body when it is located atx = 6.38 × 10⁶ m?

F(x) = kx⁻² = -3.98 × 10¹⁵ Jm / ( 6.38 × 10⁶ m)²

The code for the Google calculator: (-3.98E15J*m) / ( 6.38E6m )^2

Google calculator's result: -97.7781272 newtons

Problem's answer: F = -97.8 N or 97.8 N towards x = 0

Physics Problem with Solution - Average Power Output

What is the average power output of an engine when a car of mass M accelerates uniformly (a=const) from rest to a final speed V over a distance d on flat, level ground?
Ignore energy lost due to friction and air resistance.
Derive a formula for average power Pₐ in terms of variables: the mass M, the final speed V, and the distance d.
Pₐ=Pₐ(M,V,d) ?

Solution:

Pₐ=W/t

K-Kₒ=W

Kₒ=0

K=½MV²

Pₐ=½MV²/t

V=at

d=½at²= ½Vt

t=2d/V

Pₐ=½MV²/t=½MV²/(2d/V) = ¼MV³/d

Problem from Quiz 5

A train with the mass M accelerates uniformly from the rest to the speed V when passing through a distance d. 

What is its kinetic energy
when it passing through a distance d₁ (d₁ < d)?

Given Data: M = 15,000 kg

Vₒ = 0

V = 15 m / s

d = 150 m

d₁ = 75 m

K₁ = ?

Solution:

K(kinetic energy) = Kₒ(kinetic energy initial) + W(work) = W

W = Fx

K = W = Fd = ½MV²

F = ½MV² / d

K₁ = W₁ = Fd₁ = ½MV² d₁ / d

K₁ = ½(15,000 kg)(15 m/s)² (75m) / (150m) = 843750 J

Google Calculator Code:

0.5*(15000 kg)*(15 m/s)^2*(75m) / (150m)

Answer: 840 kJ

Precision

#Physics #PhysicsProblem #PhysicalProblem #GeneralPhysics #PhysicsI #Physics1 #CalculusBasedPhysics #CalculusBasedPhysicsI #CalculusBasedPhysics1

A photo posted by Vasiliy S. Znamenskiy (@znamenski) on Sep 28, 2016 at 12:42pm PDT

Average Power

7.36. The electric motor of a 2-kg train accelerates the train from rest to 1 m/s in 1 s. Find the average power delivered to the train during the acceleration.

Solution

Given Data	Useful formulas
M=2kg V₀=0 V=1m/s t=1s pₐ=?	K=½MV² K₀=0 pₐ=W/t W= K-K₀	pₐ=W/t=(K-K₀)/t=½MV²/t pₐ = ½MV²/t pₐ= ½ (2kg) (1m/s)² /1s = =1 kg m/s²∙m/s=1N∙m/s=1J/s=1W pₐ=1W

7.40. A 1000-kg elevator starts from rest. It moves upward for 4 s with constant acceleration until it reaches its cruising speed of 2 m/s.

(a) What is the average power of the elevator motor during this period?

(b) What is the motor power when the elevator moves at its cruising speed?

Given Data	Useful formulas
M=1000kg (a) V₀=0 V=2m/s t=4s pₐ=? (b) V=const=2m/s	K=½MV² K₀=0 U=Mgh pₐ=W/t W= (K+U)-(K₀+ U₀) h=V₀t+at²/2 a=(V-V₀)/t p = F V = MgV	W= (K+U)-(K₀+ U₀)= K+U=½MV²+Mgh=½MV²+Mgh= h=at²/2=(V/t)∙t²/2=Vt/2 W= ½MV²+MgVt/2 pₐ=W/t=½MV²/t+MgV/2 pₐ=W/t=M(V/t+g)∙½V ½MV²/t=½(1000kg)(2m/s)²/4s=500W MgV/2=1000kg∙10m/s²∙2m/s /2=10000W pₐ=10500W=10.5kW or pₐ=1000kg ∙ (2 m/s / 4 s + 10 m/s²) ∙½ ∙ 2 m/s = 10.5kW p=MgV=1000kg∙10m/s²∙2m/s=20000W=20kW

Deadline for Online Quizzes 2,3,4,5, and 6

The announcement for CityTech students: The deadline for online quizzes 2,3,4,5, and 6 is October 6 (Thursday), 4:00 PM.

After the deadline, these quizzes will be closed for records, and correct answers will be displayed.

What is the displacement of the car between t₁ = 1.0 s and t₂ = 2.0 s?

A car moving in the x direction has acceleration aₓ, that varies with time as shown in the figure. pic.twitter.com/xLnJUtQIyp

— Physics I (@PHY210) September 28, 2016

Problem from Online Quiz 1

The measured length of a cylindrical laser beam is 12.1 meters, its measured diameter is 0.00121 meters, and its measured intensity is 1.21 × 10⁵ W/m² . Which of these measurements is the most precise?

Solution:
Uncertainty of the measurement result of 12.1 m is 0.05 m so the precision can be shown by the relative error of 0.05m / 12.1m = 0.05/12.1

Uncertainty of the measurement result of 0.00121 m is 0.000005 m so the precision can be shown by the relative error of 0.000005m / 0.00121m = 0.05/12.1

Uncertainty of the measurement result of 1.21 × 10⁵ W/m² is 0.05 × 10⁵ W/m² so the precision can be shown by the relative error of (1.21 × 10⁵ W/m² ) / (0.05 × 10⁵ W/m²) = 0.05/12.1

As relative errors are the same then precision levels are the same.

PHYS 1441 Quiz 1 (eztestonline.com/695230/14564410281317700840.tp4) is closed for records. Nothing recorded.

PHYS 1441 Quiz 1 (eztestonline.com/695230/14564410281317700840.tp4) is closed for records. Nothing recorded.

Quiz 1 is closed for records. Nothing recorded.

PHY 215 Quiz 1 (eztestonline.com/695230/14274823527240100.tp4) is closed for records. Nothing recorded.

A rocket, speeding along toward Alpha Centauri, has an acceleration a(t) = At²...

^{A rocket, speeding along toward Alpha Centauri, has an acceleration a(t) = At². Assume that the rocket began at rest at the Earth (x = 0) at t = 0. Assuming it simply travels in a straight line from Earth to Alpha Centauri (and beyond), what is the ratio of the speed of the rocket when it has covered half the distance to the star to its speed when it has traveled half the time necessary to reach Alpha Centauri?}

^{a(t) = At²}

^{v(0) = 0}

^{x(0) = 0}

^{D = x(T)}

^{D / 2 = x(τ)}

^{v(τ) / v(½T) = ?}

^Solution

^{a(t) = At²}

^{v(t) = ⅓At³}

^{x(t) = At⁴ / 12}

^{D = AT⁴ / 12}

^{T⁴ = 12D / A}

^{t⁴ = 12x / A}

^{τ⁴ = 12(D / 2) / A}

^{τ⁴ = 12D / A · (½) = (½)T⁴}

^{τ = ⁴√(½) · T}

^{x(τ) = Aτ⁴ / 12 = A(⁴√(½) · T)⁴ / 12 = ½ AT⁴ / 12 = ½D}

^{v(½T) = ⅓A(½T)³ = (½)³ · ⅓AT³}

^{v(τ) = v(⁴√(½) · T) = ⅓A(⁴√(½) · T)³ = (⁴√(½))³ · ⅓AT³}

^{v(τ) / v(½T) = {(⁴√(½))³ · ⅓AT³} / {(½)³ · ⅓AT³} = {(⁴√(½))³ } / {(½)³} = (2 / ⁴√2)³ = 8 / ⁴√8 = (⁴√8)³}

Space station gives physics a boost - Bad Astronomy

Space station gives physics a boost - Bad Astronomy: This is one of the coolest videos I’ve seen in a while: during a routine reboost of the International Space Station to a higher orbit, the astronauts on board show that the station tries to leave them behind! What a fantastic example of Newtons’s First law: an object in motion tends to stay in motion …

Problem: The figure shows the position of a car (black circles) at one-second intervals. What is the acceleration at the time t = 4 s

Problem

The figure shows the position of a car (black circles) at one-second intervals.
What is the acceleration at the time t = 4 s?
 

Solution 1:

x(3s) = 24m

x(4s) = 33m

x(5s) = 40m

v(3.5s) ≈ {x(4s) - x(3s)} / 1s ≈ {3m - 24m} / 1s ≈ 9m / s

v(4.5s) ≈ {x(5s) - x(4s)} / 1s ≈ {40m - 33m} / 1s ≈ 7m / s

a(4s) ≈ {v(4.5s) - v(3.5s)} / 1s ≈ {7m / s - 9m / s} / 1s ≈ { - 2m / s} / 1s ≈ - 2m / s²

Solution 2:

a(t) ≈ {v(t + ½Δt) - v(t - ½Δt)} / Δt

v(t) ≈ {x(t + ½Δt) - x(t - ½Δt)} / Δt

v(t + ½Δt) ≈ {x(t+Δt) - x(t)} / Δt

v(t - ½Δt) ≈ {x(t) - x(t-Δt )} / Δt

a(t) ≈ ([{x(t+Δt) - x(t)} / Δt] - [{x(t) - x(t-Δt )} / Δt]) / Δt

a(t) ≈ ({x(t+Δt) - x(t)} - {x(t) - x(t-Δt )}) / (Δt)²

a(t) ≈ {x(t+Δt) - 2x(t) + x(t-Δt )} / (Δt)² derived important formula

Δt = 1s

t = 4s

a(4s) ≈ {x(5s) - 2x(4s) + x(3s)} / (1s)²

a(4s) ≈ {40m - 2 ∙ 33m + 24m} / 1s² ≈ {40m - 66m + 24m} / 1s² ≈ - 2m / 1s² ≈ - 2m / s²

Solution 3:

if a = const

x(t+Δt) = x(t) + v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²

x(t-Δt) = x(t) - v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²

x(t+Δt) + x(t-Δt) = [x(t) + v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²] + [x(t) - v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²]

x(t+Δt) + x(t-Δt) = 2x(t) + a ∙ (Δt)²

x(t+Δt) + x(t-Δt) - 2x(t) = a ∙ (Δt)²

x(t+Δt) - 2x(t) + x(t-Δt) = a ∙ (Δt)²

a = {x(t+Δt) - 2x(t) + x(t-Δt)} / (Δt)² derived important formula

Δt = 1s

t = 4s
a = {x(5s) - 2x(4s) + x(3s)} / (1s)²

a = {40m - 2 ∙ 33m + 24m} / 1s² = {40m - 66m + 24m} / 1s² = - 2m / 1s² = - 2m / s²

2 Describing Motion: Kinematics in One Dimension

For students PHYS 1441 (NYC College of Technology) E474 [54392]: pic.twitter.com/Ta5cpIU7vK

— Physics I (@PHY210) September 1, 2016

PDF Versions of Textbook

In general, all exams in my class are open-book exams. It means students may use paper textbooks during exams. But I don't give permission to use in exams of any electronic devices except simple calculators. Students may use PDF versions of the textbook during lectures, laboratory lessons, or at home.