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A girl is sitting in a sled sliding horizontally along some snow (there is friction present).

The normal force supporting the girl and sled is 636 N.

If the coefficient of static friction is 0.38, and the coefficient of kinetic friction between snow and sled is 0.25, what is the force of friction exerted on the sled?

Answer :

opudodennis

Answer:

159 N

Explanation:

The force of friction, Fr is a product of coefficient of feiction and the normal force. Therefore, Fr=uN where N is the normal force and u is coefficient of friction. Here, we have two coefficients of friction but since it is sliding, then we use coefficient of kinetic energy. Substituting 0.25 for u and 636 N for N then

Fr=0.25*636=159 N

Therefore, the force of friction is equivalent to 159 N

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