### Forces on Inclined Planes:Problem:

**Problem:**

A skier goes down a smooth 30o hill (frictionless) for a distance of 10 m to the bottom of the hill where he then continues on a frozen, frictionless pond. After that, he goes up a hill inclined at 25o to the horizontal. How far up this second hill does he go before stopping if the coefficient of friction on this hill is 0.10?

**Analysis and Solution:**

**Part I - Accelerated Motion**

a = (g)(sin30^{o})

= 9.8 X 0.5 m/s^{2}

V1 = 0 ( V_{2} ) ^{2 }= ( V_{2} ) 1 + 2(a)(d) d = 10 m

( V2 ) = 9.9 m/s

**Part II - No acceleration**

No acceleation means a = 0, therefore, Fnet = ma = 0 and V_{1} is the same as V_{2} from Part 1.

**Part III - Friction up an inclined plane**

From the Free Body Diagram above

**Fnet = -Fgx - Ff **

ma = -(m)(g)(sin 25^{o}) - µ(m)(g)(cos 25^{o})

**m drops off on both sides of the equation**

**This is proof that the acceleration of an objevct sliding down an inclined plane is independent of the mass of the object.**

What factors influence the acceleration of the object?

Hint: Look at the above equation.

we solve for the acceleration, a

a = - 4.99 m/s^{2}

we use the formula below and the acceleration found above to find the distance, d

Where V_{2} is zero (that's when the skier stops)

Solving for d, where V1 = 9.9 m/s from part 1, and a = -4.99 m/s^{2}

We obtain a value for d:

d = 9.8 m

Therefore the distance traveled by the skier before he stops up the second hill is 9.8 m