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A cube of mass m = 0.49 kg is set against a spring with a spring constant of k1 = 606 N/m which has been compressed by a distance of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 = 233 N/m. How far d2, in meters, will the second spring compress when the block runs into it?

Respuesta :

Answer:0.161 m

Explanation:

Given

mass of cube  [tex]m=0.49\ kg[/tex]

Spring constant [tex]k_1=606\ N/m[/tex]

compression in the spring [tex]x_1=0.1\ m[/tex]

When this cube is released then it will compress another spring of spring constant [tex]k_2=233\ N/m[/tex]

Conserving energy

[tex]\frac{1}{2}k_1x^2=\frac{1}{2}mv^2=\frac{1}{2}k_2x'^2[/tex]

[tex]\frac{x'}{x}=\sqrt{\frac{k_1}{k_2}}[/tex]

[tex]x'=0.1\times \sqrt{\frac{606}{233}}[/tex]

[tex]x'=0.1\times 1.61[/tex]

[tex]x'=0.161\ m[/tex]

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