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Яндекс.Метрика

Wednesday, October 12, 2016

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 - U
U₂ = U₁ + mg(y₂ - y₁)

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

Problem's answer is 23 J.



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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.
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

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 at
x = 6.38 × 10⁶ m?

F(x) = kx⁻² = -3.98 × 10¹⁵ Jm / ( 6.38 × 10 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 P in terms of variables: the mass M, the final speed V, and the distance d.
P
=P(M,V,d) ?

Solution:
Pₐ=W/t
K-Kₒ=W
Kₒ=0
KMV²
PₐMV²/t
V=at
d=½at²= ½Vt
t=2d/V


Pₐ=½MV²/t=½MV²/(2d/V) = ¼MV³/d