Question

A skier of mass 75kg starts from rest at the top of a friction less incline that is 20.0 m in height. As soon as she touches the bottom of the incline, she encounters a horizontal surface that is 1000 m long and the skier eventually comes to rest. The coefficient of kinetic friction between skier and snow is 0.205. Find the distance the skier covers before coming to rest. Show all work.

Answer 1

Here, let's break this question into two parts

Part 1 = when a skier was on top and travelled on an inclined plane without friction. from this part, we have to find the velocity of skier after reaching the end of the inclined plane

As the energy is conserved because the surface is frictionless

Potential energy = kinetic energy

P.E = mass * g * height

P.E = 75*9.81*20 = 14700J

Now, K.E = 1/2*m*v^2

K.E = 0.5*75*v^2 = 37.5*v^2 J

As P.E = K.E

14700 = 37.5* v^2

v = Sqrt 392 = 19.7m/s

Hence when the skier is at end of the inclined plane he has the velocity of 19.7 m/s

Part 2

Now skier is at horizontal plane now we have to find distance covered by him until he came to rest before that, we have to find acceleration when there is a dynamic friction, as skier moving on a horizontal surface the only force acting is the frictional force opposite to the motion

Fr = mg * 0.205 = 75 * 9.8 *0.205 = 150.6N

This force is opposite to motion force

m*a = -Fr

a = -Fr/m = 150.6/75 = -2 m/s^2

Now, using the equation of motion

u^2 = v^2 + 2*a*s

Where u = final velocity = 0 because skier come to rest , v = initial velocity =19.7 , s = distance travelled

0 = 19.7^2 + 2*(-2)*s

388 = 4*s

s = 388/4 = 97m

Hence, the distance travelled by skier is 97m before coming to rest