A 20-kg block slides down a fixed rough curved track The block has a speed of 5 0 m/s after its height above a horizontal surface has decreased by 1.8 m. Assume the block starts from rest. How much work is done on the block by the force of friction during this descent?​

U = 102.8 J (100 J to two significant digits)

Explanation:

potential energy converted = 20(9.8)(1.8) = 352.8 J

kinetic energy at base of track = ½(20)5.0² = 250 J

energy (work) of friction 352.8 - 250 = 102.8 J

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