**Unformatted text preview: **f the distance x it is stretched. Take the zero of potential energy in the unstretehed position. 18. A 3.D-kgfish is hangi~g from a spring scale whose spring constant is 240 N/m. (a) What is the potential energy of the spring? (b) If the fish were moved slowly upward to the equilibrium position of the spring, by how much would its gravitational potential energy change? (c) In case (b), by how much would the spring's potential energy change? Explain ariy apparent discrepancies. Solution
Assuming that the direction of F is opposite to. the displacement, we find: U=- r Jo F.dx'=FoJo
Folx' + r [i+X' -f.-- (f+x')2 2] dx' Solution
To avoid confusion, call the place where the fish hangs at rest the equilibrium position (B), and the place = ~> (l:x') [ = Fo [X + ~; + (f ~ X) - . CHAPTER 8 127 roblem is identical to Problem 7-33. x' is a yvariable.) Solution
If the slope is frictionless (and there are no other losses of energy), the total mechanical energy of the spring and mass is conserved. Initially. Ko = 0, and Uo == ~ kx2 + mgyo (where we neglect the gravitational potential energy of the spring). Then Equation 8-8 gives U...

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