Answered

A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 72.0 J. What compression (as measured from equilibrium) would result in 100 J being stored in this case?

Answer :

Answer:14.14 cm

Explanation:

Given

Spring Compression [tex]x=12 cm[/tex]

Potential energy Stored in spring[tex]=72 J[/tex]

Suppose k is the spring constant of spring

Potential Energy of spring is given by [tex]=\frac{kx^2}{2}[/tex]

[tex]\frac{k(0.12)^2}{2}=72[/tex]

[tex]k(0.12)^2=144[/tex]

[tex]k=10,000 N/m[/tex]

[tex]k=10 kN/m[/tex]

for 100 J energy

[tex]\frac{k(x_0)^2}{2}=100[/tex]

[tex]10\times 10^3\cdot (x_0)^2=200[/tex]

[tex](x_0)^2=2\times 10^{-2}[/tex]

[tex]x_0=0.1414[/tex]

[tex]x_0=14.14 cm[/tex]

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