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A spring stores potential energy U0 when it is compressed a distance x0 from its uncompressed length.

a] In terms of U0, how much energy does it store when it is compressed twice as much? Uel/U0 =?

b] In terms of U0, how much energy does it store when it is compressed half as much? Uel/U0 = ?

c] In terms of x0, how much must it be compressed from its uncompressed length to store twice as much energy? x/x0 = ?

d] In terms of x0, how much must it be compressed from its uncompressed length to store half as much energy? x/x0 = ?

Respuesta :

Answer:

Explanation:

Using Hooke's law

Elastic potential energy = 1/2 K x²

K is elastic constant of the spring

x is the extension of the spring

a) The elastic potential energy when the spring is compressed twice as much  Uel = 1/2 k (2x₀) ² =  4 (1/2 kx₀²)= 4 U₀

b) when is compressed half as much Uel = 1/2 k [tex]\frac{x0}{4} ^{2}[/tex] = [tex]\frac{1}{4}[/tex] ( U₀)

c) make x₀ subject of the formula in the equation for elastic potential

      x₀ =[tex]\sqrt{\frac{2U0}{K} }[/tex]

x, the amount it will compressed to tore twice as much energy = [tex]\sqrt{\frac{2 (2U0)}{K} }[/tex]

x = √2 x₀

d) x₁, the new length it must be compressed to store half as much energy = [tex]\sqrt{\frac{2 (\frac{1}{2})U0 }{K} }[/tex]

x₁ = [tex]\sqrt{\frac{1}{2} }[/tex] x₀

batolisis

A) The amount of energy stored when compressed twice as much is : 4 U₀

B) The amount of energy stored when compressed half as much is : [tex]\frac{1}{4} ( U_{o})[/tex]

C) The amount of compression from its uncompressed length to store twice as much energy is : x = √2 x₀

D) The amount of compression from its uncompressed length to store half as much energy is : x₁ = [tex]\sqrt{\frac{1}{2} }[/tex] x₀

Determine the amount of energy and the compression

A) The amount of energy stored when compressed twice

Uel / Uo =

Uel = [tex]\frac{1}{2}[/tex] k (2x₀) ² =  4 ([tex]\frac{1}{2}[/tex] kx₀²) = 4 U₀

Where : K = elastic constant, x = string extension,

B) The amount of energy stored when compressed half as much

Uel/U0 =

1/2 k [tex]\frac{x_{o}^2 }{4}[/tex] = 1/4 ( U₀)

Where : Elastic potential energy = [tex]\frac{1}{2}[/tex] Kx²

C) The amount of compression from its uncompressed length to store twice as much energy

x / xo =

we will use the expression below

Xo = [tex]\sqrt{\frac{2Uo}{k} }[/tex]

where ; x = [tex]\sqrt{\frac{2(2Uo)}{k} }[/tex]

therefore ; x / xo = [tex]\sqrt{2} x_{o}[/tex]

D) The amount  ( X₁ ) it must be compressed to store half as much energy can be expressed as : X₁ = [tex]\sqrt{\frac{1}{2} }[/tex]  x₀

Hence we can conclude that the answers to your questions are as listed above

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