A skier of mass 55.0 kg slides down a slope 11.7 m long, inclined at an angle theta to the horizontal. The magnitude of the kinetic friction...

The skier slides down a slope that is 11.7 m long and
inclined at an angle theta to the horizontal. This gives the height of the skier at the
start as 11.7*sin theta.

The potential energy of the skier
here is m*g*h = 55*11.7*sin theta*9.8. The initial speed of the skier is given as 65.7
cm/s of 0.657 m/s. Kinetic energy is (1/2)*m*v^2 =
(1/2)*55*(0.657)^2

The total energy is 55*11.7*sin
theta*9.8 + (1/2)*55*(0.657)^2

At the bottom of the slope
the skier's speed is 7.19 m/s, the kinetic energy is (1/2)*55*(7.19)^2. The potential
energy is 0.

This gives total energy as
(1/2)*55*(7.19)^2.

The force of resistance due to friction
is 45.1 N

From the conservation of
energy:

55*11.7*sin theta*9.8 + (1/2)*55*(0.657)^2 - 45.1 =
(1/2)*55*(7.19)^2

=> 55*11.7*sin theta*9.8 =
(1/2)*55*(7.19)^2 + 45.1 - (1/2)*55*(0.657)^2

=>
55*11.7*sin theta*9.8 = 1454.872

=> sin theta =
0.2307

=> theta = arc sin
(0.2228)

=> theta = 13.3338
degrees.

The required slope of the incline is
13.3338 degrees.