Question

90 kgkg ice skater moving at 10.0 m/sm/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of 0.410. How far along the rough ice does she go before stopping?

Answer 1

Answer:

12.43m

Explanation:

information we know:

mass:

$m=90kg$

initial velocity:

$v_{i}=10m/s$

coefficient of kinetic friction:

$\mu=0.41$

with this information we use Newton's second law:

$F=ma$

where $F$ is the force acting on the person, and $a$ is its acceleration.

analyzing the forces acting the vertical axis y:

$N-mg=ma$

where N is the normal force and -mg is the weight of the person (weight and normal force are the only two forces acting on the vertical axis)

since the person is not moving up or down there is no acceleration in the y axis (a=0):

$N-mg=0\\N=mg$

and now analyzing the forces acting on the horizontal axis x (using Newton's second law):

$F=ma$

the only horizontal force on the person is the fricction, so:

$-F_{fr}=ma$

where $-F_{fr}$ is the force due to friction (negative because it opposes the movement) and it is defined as:

$F_{fr}=\mu N$

and since we know that N=mg:

$F_{fr}=\mu mg$

thus the second law for the horizontal movement is:

$-\mu mg=ma\\-\mu g=a$

substituting $\mu=0.41$ and  $g=9.81m/s^2$

$a=-(0.41)(9.81m/s^2)\\a=-4.0221m/s^2$

and then we use the following  kinematic equation :

$v_{f}^2=v_{i}^2+2ax$

to find the displacement x of the ice skater, since the final velocity is equal to zero  (vf=0 , because the ice skaters stops), we get

$0=v_{i}^2+2ax\\v_{i}^2=-2ax\\\frac{v_{i}^2}{-2a}=x$

and we substitute the known values:

$\frac{(10m/s)^2}{-2(-4.0221m/s^2)}=x\\ 12.43m=x$

the answer is 12.43m