Two identical springs with spring constant k are connected to identical masses

Two identical #springs with spring #constant k are #connected to #identical #masses of #mass M, as shown in the figures. #Physics #problem pic.twitter.com/8ff4nK5pcn

— Vasiliy S Znamenskiy (@znamenski) October 24, 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

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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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