Question

An ice cube with a mass of 0.0580 kg is placed at the midpoint of a 1.00-m-long wooden board that is propped up at a 46° angle. The coefficient of kinetic friction between the ice and the wood is 0.205.

(a) How much time does it take for the ice cube to slide to the lower end of the board?
  s

(b) If the ice cube is replaced with a 0.0580-kg wooden block, where the coefficient of kinetic friction between the block and the board is 0.325, at what angle should the board be placed so that the block takes the same amount of time to slide to the lower end as the ice cube does? You may find a spreadsheet program helpful in answering this question.

Answer 1

Baloncing forces mg sino - fama f=MF = Mongcosa a mesino o mo ng con so a a= o sino - Mgcaso = 9.8x sinu 6-0.205x9.8x cos46 = 5.65 m/s² distance travelled d=tat? t = /22 = 2x0.5 5.65 t = 0.42s For some amount of time acceleration should be some a = 5.65 5.65 =9.8( sino-0.325 caso) 0.576 = sino-0.325 Si- sinza on squaring 0.33 + sino - 2x0.576xsino = (0.325² (1-sino) 0.33 + sinzo - 1.152 sino = 0.1056 - 0-1056 sino 220 -1-152 sino to.2244 20 a 온우나나 s sino=0260.78 o= 15.07 51.5° 1.1056 o=51.5°

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