Question

A skier moving at 5.30 m/s encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. How far does she travel on this patch before stopping?

Answer 1

Answer:

6.5 m

Explanation:

First of all, we need to compute the acceleration of the skier. We know that there is only force acting on the skier: the force of friction, which is in the opposite direction to the motion of the skier. Using 2nd Newton Law:

$\sum F = ma\\F_f = ma\\-\mu m g = ma$

where $F_f = -\mu mg$ is the force of friction, with

$\mu=0.220$ is the coefficient of kinetic friction

m is the mass of the skier

$g=9.8 m/s^2$ is the acceleration due to gravity

Re-arranging the equation, we find:

$a=\mu g=-(0.220)(9.8 m/s^2)=-2.16 m/s^2$

So now we can use the following SUVAT equation to calcualte the total displacement of the skier before stopping:

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity

u = 5.30 m/s is the initial velocity

a = -2.16 m/s^2 is the acceleration

d = ? is the displacement

Solving the formula for d, we find:

$d=\frac{v^2-u^2}{2a}=\frac{0-(5.30 m/s)^2}{2(-2.16 m/s^2)}=6.5 m$