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

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

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.

Problem and Solution. 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}

Problem: The figure shows the position of a car (black circles) at one-second intervals...

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