A loaded 335 kg toboggan is traveling on smooth horizontal snow at 4.60 m/s when

Question

A loaded 335 kg toboggan is traveling on smooth horizontal snow at 4.60 m/s when it suddenly comes to a rough region. The region is 8.30m long and reduces the toboggan's speed by 1.30m/s . What average friction force did the rough region exert on the toboggan?By what percent did the rough region reduce the toboggan's kinetic energy and speed?

Explanation / Answer

v1= 4.6m/s v2=4.6-1.3=3.3m/s m=335kg initial energy= initial kinetic energy = (m* v1*v1 )/2=3544.2 joules. final energy=work done by friction + final kinetic energy = work done by friction + 1824.075 Joules. by conservation of energy :: initial energy = final energy => work done by friction= average friction * distance =3544.2 - 1824.075 =1720.125 joules therefore, average frictional force= 1720.125/8.3 = 207.25 N. % reduction in kinetic energy = (initial kinetic energy - final kinetic energy) / initial kinetic energy. =0.485 =48.5%..

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