Unformatted text preview: 1 + 2 Î» R 2 k (c) We seek the conditions under which V ( Î» ) = 2 V (0) . Using the result of Part (b), we find 2 R 5 k m 5 1 + 2 Î» R 2 k = 4 R 5 k m = â‡’ 1 + 2 Î» R 2 k = 4 Solving for Î» , we conclude that Î» = 3 k 2 R 2...
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 Spring '06
 Shiflett
 Energy, Potential Energy, 3k, 4 k, initial kinetic energy, Principle of Work and Energy

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