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 23 Click here to jump to the problem! Ryry0132019-05-26 15:30:53 Slightly more accurate would be:\r\n\r\nEnergy before: $mgh + \\frac{1}{2} mv^{2}$\r\nEnergy after: $0 + \\frac{1}{2} mv^{2}$ (same velocity throughout)\r\n\r\nThen, the mgh-->0 part all went into friction, as you all said.Click here to jump to the problem!