222222112 21212()11122()22iimvKm m vmm+Δ=+−−=−−+which yields ΔK= –35 J. (Although it is not necessary to do so, still it is worth noting that algebraic manipulation of the above expression leads to ΔKv=+12direl2where vrel= v1– v2). Conservation of energy then requires 2122( 35 J)21120 N/mKkxKxk−Δ− −=−Δ¡=== 0.25 m. 59. As hinted in the problem statement, the velocity
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