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A spring whose stiffness is 1140 n/m has a relaxed length of 0.51 m. if the length of the spring changes from 0.26 m to 0.79 m, what is the change in the potential energy of the spring? δu = -9.063 incorrect: your answer is incorrect. j

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A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs. 
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The change in potential energy of the spring will be =332.9 J

What is potential energy?

The potential energy is defined as the energy possessed by any object due to their position with respect to thier mean position.

The energy of the spring is given as

[tex]E=\dfrac{1}{2}kx^2[/tex]

Now PE at point x= 0.26 m

[tex]E_1=\dfrac{1}{2}\times 1140\times 0.26^2=22.8\ J[/tex]

Now PE at point x= 0.79 m

[tex]E_2=\dfrac{1}{2}\times 1140\times (0.79)^2=355.7\ J[/tex]

The change in potential energy will be

[tex]\Delta E=E_2-E_1=355.7-22.8=332.9\ J[/tex]

Hence the change in potential energy of the spring will be =332.9 J

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