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

How much gravitational potential energy does the object have?

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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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.

How much does the spring compress?

A 2.0 - kg mass is dropped 2.0 m above a spring with a spring constant 40.0 N/m. How much does the spring compress? Use g = 10 m/s².

Solution

Given Data:
m = 2.0kg
k = 40.0N/m
h = 2.0m
x = ?

Useful formulas:
Uₛ = ½kx²
Uₘ = mgh

Solution:
mg(h + x) = ½kx²
½kx² - mgx – mgh = 0

x² - (2mg/k)x – (2mg/k)h = 0

x² – 2 (mg/k)x + (mg/k)² = (mg/k)² + (2mg/k)h

(x - mg/k)² = (mg/k)² + (2mg/k)h

x - mg/k = ±√ {(mg/k)² + 2(mg/k)h}

x = mg/k±√ {(mg/k)² + 2(mg/k)h}

Calculation
mg/k = 2kg*10m/s ² / 40N/m = ½m

(mg/k)² = ¼m²

2(mg/k)h = 2·½m·2m = 2m²

(mg/k)² + 2(mg/k)h = ¼m² + 2m² = (⁹/₄)m²

√ {(mg/k)² + 2(mg/k)h} = (³/₂)m

x₁ = ½m + (³/₂)m = 2m
x₂ = - ½m - (³/₂)m = - 1m

Problem's answer: 2.0 m

Physics Problem - Potential Energy

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

How Much Work Is Done by the Spring on the Mass?

A horizontal spring having constant k is attached to a block having mass M. The mass is oscillating freely on a frictionless table with amplitude A. As the mass travels from the position of the maximum stretch to half the position of maximum stretch, how much work is done by the spring on the mass?

Given Data

k = 50.0 N/m

M= 0.50 kg

A= 15.0 cm

xₒ=A

x=A/2

W=?

Solution

U=½kx²

Uₒ=½kxₒ²

Uₒ-U=W

W=½kxₒ²-½kx²=½k(xₒ²-x²)

W=½k(A²-(½A)²)=½k(A²-(½A)²)=½k(A²-¼A²)=½k(¾A²)=⅜kA²

W=⅜·(50 N/m)·(0.15m)²

Code for the Google Calculator: 3/8*(50 N/m)*(0.15m)^2

Google Calculator's result: 0.421875 joules

Answer:   W=0.42J

Quiz 5, 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 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) ?

Pₐ=W/t

K-Kₒ=W

Kₒ=0

K=½MV²

Pₐ=½MV²/t

V=at

d=½at²= ½Vt

t=2d/V

Pₐ=½MV²/(2d/V) = ¼MV³/d