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A 1.1 g marble is fired vertically upward using a spring gun. The spring must be compressed 6.2 cm if the marble is to just reach a target 16 m above the marble's position on the compressed spring. (a) What is the change ΔUg in the gravitational potential energy of the marble-Earth system during the 16 m ascent

Respuesta :

jolis1796

Answer:

ΔU = 0.17 J

Explanation:

m = 1.1 g = 1.1*10⁻³ Kg

h₁ = 6.2 cm = 0.062 m

h₂ = 16.062 m

ΔU = ?

We apply the formulas

U₁ = m*g*h₁

U₂ = m*g*h₂

ΔU = U₂ - U₁ = m*g*h₂ - m*g*h₁ = m*h*(h₂ - h₁)

⇒  ΔU = (1.1*10⁻³ Kg)*(9.81 m/s²)*(16.062 m - 0.062m)

⇒  ΔU =  0.17 J

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