A tobogganer coasts down a hill and then goes over a slightrise with speed 1.5 m

Question

A tobogganer coasts down a hill and then goes over a slightrise with speed 1.5 m/s. The top of this rise can be taken to be acircle of radius 3 m. The toboggan and occupant have a combinedmass of 110 kg. If the coefficient of kinetic friction between thesnow and the toboggan is 0.10, what frictional force is exerted onthe toboggan by the snow as the toboggan goes over the top of therise? How do I set this problem up and complete it? A tobogganer coasts down a hill and then goes over a slightrise with speed 1.5 m/s. The top of this rise can be taken to be acircle of radius 3 m. The toboggan and occupant have a combinedmass of 110 kg. If the coefficient of kinetic friction between thesnow and the toboggan is 0.10, what frictional force is exerted onthe toboggan by the snow as the toboggan goes over the top of therise? How do I set this problem up and complete it?

Explanation / Answer

The normal force acting on the toboggan and occupant     N = mg - mv2 / r                         = 110 * 9.8 - 110*1.52 / 3                         = 995.5 N The frictional force acting toboggan is               Ffric = N                       = 0.10 * 995.5                       = 99.55 N

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