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

6.5 m

Step-by-step 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=(v^2-u^2)/(2a)=(0-(5.30 m/s)^2)/(2(-2.16 m/s^2))=6.5 m`