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℀℁℅℆
Wednesday, October 12, 2016
Friday, October 7, 2016
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.
Tuesday, October 4, 2016
Physics Problem - A spring is fixed vertically to the floor
A #spring is fixed #vertically to the floor.
— Physics I (@PHY210) October 5, 2016
A 10-kg #box is then #pressed down such that the spring is #compressed https://t.co/Z5fLcbq2LX
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
Sunday, October 2, 2016
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?
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
Saturday, October 1, 2016
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) ?
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
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