A 20kg piece of wood slides on the surface shown in Fig P747

A 2.0-kg piece of wood slides on the surface shown in Fig. P7.47. The curved sides are perfectly smooth, but the rough horizontal bottom is 30 m long and has a kinetic friction coefficient of 0.20 with the wood. The piece of wood starts from rest 4.0 m above the rough bottom. (a) Where will this wood eventually come to rest? (b) For the motion from the initial release until the piece of wood comes to rest, what is the total amount of work done by friction?

Please show work step by step, thank you!

Solution

GPE = m g h = 2.0 * 9.8 * 4 = 78.4 J [total mechanical energy ]

friction force = (mu_k x Normal force) = 0.20 m g

= 0.20 * 2.0 * 9.8 = 3.92 N

3.92 * d = 78.4

d = 20 m

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Total amount of friction work must equal block\'s total ME

= m g h = 2.0 * 9.8 * 4 = 78.4 J