Question

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 1

159 N

Step-by-step explanation:

There are two type of forces of friction:

- Force of static friction: this force occurs when an object is at rest over a surface - if we try to push the object to put it in motion, we observe that there is a resistive force that prevents the object from moving - this force is the force of static friction.

- Force of kinetic friction: this force is a resistive force that occurs when the object is already in motion - it opposes the motion of the object, being in the direction opposite to the motion

In this problem, the sled with the girl is sliding along the snow: so it is in motion, so we want to calculate the force of kinetic friction.

This force is given by:

`F_f = \mu_k N`

where

`\mu_k` is the coefficient of kinetic friction

N is the normal force on the system

In this problem,

`\mu_k = 0.25`

N = 636 N

So the force of friction is

`F_f=(0.25)(636)=159 N`