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Elongation and potential energy of two springs

Spring S has a relaxed length of s= .85 m and has a force constant ks= 24 nt/m. Spring P has a relaxed length p= .45 m and has a force constant ks= 60 nt/m. Initially, each spring has one end attached to opposite sides of a box so that the distance between their free ends is d= .70 m apart. Their ends are pulled toward each other and joined.
See ATTACHMENT for a diagram showing parameters and physics statements.
a. Find the final length Ls and Lp of each of the joined springs.
b. Find the final amount of potential energy in each spring.

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Solution:
Step 1.
Let xs be the elongation of spring S. Then since the total sum of elongations of both springs is d, the elongation of spring P must be d - xs. By statement C on ATTACHMENT1, using Hooke's Law, we can ...

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