Chapter 13. Fluids. Giancoli, Douglas C. Physics for Scientists and Engineers with Modern Physics - 4th Ed

Gravitational Potential Energy

#PhysicsProblem #Physics #problem #object #floor #gravitational #potential #energy #Howmuch #PotentialEnergy #AboveTheFloor #GivenData pic.twitter.com/JPMwGoTW69

— Physics I (@PHY210) October 12, 2016

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.

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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

The potential energy of a 10-kg mass is given by ...

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

Constant Power

What is the power output of an engine when a car of mass M accelerates with the constant power 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 the power P in terms of variables: the mass M, the final speed V, and the distance d.
P=P(M,V,d) ?

v=v(t) The speed is a function of time.

Pt = ½m v(t)² Because of the energy conservation law.

v(t)²=2tP/m=(2P/m)t

v(t)=(2P/m)^½t^½

x(t)=(2P/m)^½ (⅔t^3/2)

x(t)²=(2P/m) (⅔)² t³

t=v(t)² / (2P/m)

t³=v(t)⁶ / (2P/m)³

x(t)²=(2P/m) (⅔)² t³=(2P/m) (⅔)² v(t)⁶ / (2P/m)³

x(t)²=(⅔)² v(t)⁶ / (2P/m) ²

x(t)=(⅔) v(t)³ / (2P/m)

2P/m = (⅔) v(t)³ / x(t)

P = ½(⅔) v(t)³m / x(t)

P = ⅓ v(t)³m / x(t) = ⅓ V³m / d

Physics Problem - Find Average Power

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... pic.twitter.com/iqz40MQpOh

— Physics I (@PHY210) October 1, 2016