A 59-kg person on skis starts from rest down a hill sloped at 36° from the horizontal. The coefficient of friction between the skis and the snow is 0.16. After the skier had been moving for 5.0 s, the friction of the snow suddenly increased and made the net force on the skier zero.

What is the new coefficient of friction?

How fast would the skier now be going after skiing for another 5.0 s?

where’s the .81 come from

To find the new coefficient of friction, we need to analyze the initial situation when the skier started moving and the situation after the friction suddenly increased.

First, let's determine the initial acceleration of the skier down the hill using Newton's second law, which states that force equals mass multiplied by acceleration (F = ma). The only force acting on the skier is the component of gravity parallel to the slope, given by F = mg sin(θ), where m is the mass in kg, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the slope.

F = (59 kg)(9.8 m/s^2) sin(36°)
F = 348.107 N

Next, we can calculate the frictional force acting on the skier using the formula F_friction = μN, where μ is the coefficient of friction and N is the normal force. The normal force is equal to the component of gravity perpendicular to the slope, given by N = mg cos(θ).

N = (59 kg)(9.8 m/s^2) cos(36°)
N = 472.543 N

F_friction = (0.16)(472.543 N)
F_friction = 75.6069 N

Since the net force on the skier is suddenly zero after 5.0 seconds, the frictional force must have increased to counterbalance the component of gravity parallel to the slope. Therefore, the new frictional force acting on the skier is equal to the initial acceleration times the mass.

F_friction = (59 kg)(a)
a = F_friction / m = 75.6069 N / 59 kg
a ≈ 1.282 m/s^2

Now, let's calculate the skier's new speed after skiing for another 5.0 seconds using the kinematic equation v = u + at, where v is the final velocity, u is the initial velocity (which is 0 m/s, since the skier starts from rest), a is the acceleration, and t is the time.

v = 0 + (1.282 m/s^2)(5.0 s)
v ≈ 6.41 m/s

Therefore, the new coefficient of friction is approximately 0.16, and the skier's speed after skiing for another 5.0 seconds is approximately 6.41 m/s.

ma=mgsinα –F(fr) = mgsinα – μmgcosα

a= g(sinα – μcosα ) = 9.8• (0.59-0.16•0.81)=4.5 m/s²
The speed after the first 5 sec is
v=at= 4.5•5=22.5 m/s
If the net force=0, a=0 =>
0 = mgsinα – μ₁•mgcosα ,
μ₁ =sinα/cosα=tanα=0.73.
If a=0 => v=const=22.5 m/s