# 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 is...45.1N. The skier's initial speed is 65.7 cm/s and...

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 is...

45.1N. The skier's initial speed is 65.7 cm/s and the speed at theĀ bottom of the slope is 7.19 m/s. Determine the angle theta from the law of conservation of energy. Air resistance is negligible.

*print*Print*list*Cite

### 2 Answers

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.**

the text book answer says 17.5, i got 12.5 degrees for my answer and i've been trying to get 17.5.