coefficient of friction?

Question

A skier traveling 12.0 m/s reaches the foot of a steady upward 18° incline and glides 10.2 m up along this slope before coming to rest. What was the average coefficient of friction

Answer 1

Draw a diagram to indicate forces on the skier perpendicular to and parallel to the slope. Forces perpendicular to the slope are the normal force N, and a component of the force due to gravity given by mg cos(18°) (where m is the skier's mass and g is the acceleration due to gravity). There is no net force perpendicular to the slope, so

N = mg cos(18°)

The skier experiences two constant forces parallel to the slope: the component of gravity parallel to the slope, of magnitude mg sin(18°); and the force due to friction, of magnitude µN (where µ is the coefficient of kinetic friction). Both forces point down the slope, opposing the skier's motion. So the net force parallel to the slope is

F = -mg sin(18°) - µN

which, using the above result for N, becomes

F = -mg sin(18°) - µmg cos(18°)

Since F = ma, where a is the acceleration, we get

a = -g sin(18°) - µg cos(18°)

To solve for µ I am going to use the equation

(v_f)² = (v_i)² + 2 a d

where v_f is the final velocity, v_i is the initial velocity, and d is the distance traveled during the time that the velocity changes from v_i to v_f. [If you don't recognize this equation, it can be obtained by solving v_f = v_i + a*t for t, then substituting for t in
d = (v_i)t + ½ a t² and solving for (v_f)²]

In this case, v_f = 0. We know d and v_i, and we have an expression for a. I leave it to you to solve for µ.

You could also solve this problem using energy considerations. At the beginning, the skier has kinetic energy given by
½ m (12.0)² . Part of this energy is converted to gravitational potential energy, and part is dissipated as heat in opposing friction.

The vertical increase in height that the skier achieves is
d sin(18°), so the gravitational potential energy is given by

mgd sin(18°)

The work done against friction is given by the friction force, times the distance through which the friction acts. This is given by

µmg cos(18°) d

where I've used the calculation obtained earlier for the friction force (so this approach is more involved than it appears).

So

½ m (12.0)² = mgd sin(18°) + µmg cos(18°) d

where d = 10.2.

The mass m divides out, and the result is readily solved for µ.

Answer 2

to discover the frictional stress, discover the stress required to realize the acceleration devoid of friction: F=m*a = 5*6 = 30 N as a results of fact you desire 40 N to realize the comparable acceleration, the friction stress could be 10 N: 40-30=10 N The coefficient of friction (µ) equals the Friction stress (Ff) divided by skill of the conventional stress (Fn). On a horizontal airplane, the conventional stress is comparable to the load: 5kg*9,8m/s^2=40 9 N µ=Ff/Fn=10N/49N=0,2