A skier  has just begun descending a 30° slope.
Assuming the coefficient of kinetic friction is 0.10, calculate:

a.Her acceleration.


b.The speed she will reach after 6.0 s.

Hi pj4444!!!

letīs work!!

Uniformly Accelerated Linear Motion equations:

X = X0 + V0 * T + 1/2 * A * t^2  [1]
Vf = V0 + A * T                  [2]
A = constant                     [3]
Vf^2 - V0^2 = 2 * A *(Xf - X0)   [4]

If Mk is the coefficient of kinetic friction, N is the Normal force
(the component force in the perpendicular direction produced by the
weight of the studied object)and Fk is the force of kinetic friction
then:
Fk = Mk * N                      [5]
F = m * A                        [6]
g = 9.81 m/s^2

Because the 30š slope, N = W * cos30    where W is the weight of the
skier.
Then 
Fk = Mk * N = 0.10 * W * cos30 = 0.10 * m * g * cos30 
then the acceleration due Fk is:
Ak = Fk / m = 0.10 * g * cos30

The skierīs acceleration due the gravity on the 30šslope is:
Ag = g * sin30

The total acceleration of the skier is:
A = Ag - Ak = g * sin30 - 0.10 * g * cos30 = 9.81 * 0.5 - 0.981 *
0.866 =
  = 4.055 m/s^2

This answer part a).


Part b).

Here we will use the equation [2]:

Vf = V0 + A * T = 0 + 4.055 * 6 = 24.33 m/s


This finish the problem.
I hope this helps you, if you need a clarification please post a
request for it.

Best regards.
livioflores-ga