# Homework Help: Conservation of energy problem(s)

1. Oct 29, 2008

### Quincy

1. The problem statement, all variables and given/known data
A ski starts from rest and slides down a 22o incline 75 m long. a) if the coefficient of friction is 0.090, what is the ski's speed at the base of the incline? b) If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level? Use energy methods.

2. Relevant equations
-Wnc = (KEf - KEo) + (PEf - PEo)

3. The attempt at a solution

a)Since the hypotenuse of the incline is 75 m, and the angle relative to the horizontal is 22o, the vertical height is 75 sin22o = 28.1 m.
PEf and KEo cancel out, and the equation becomes:

PEo = KEf + Wnc

That simplifies to:

mgh = (1/2)mv2 + umgcos22o -- (u is coefficient of friction)

masses cancel out

(9.8)(28.1) = (1/2)v2 + (0.09)(9.8)cos22o

v = 23.4 m/s -- According to the book, the answer is 21 m/s. What am I doing wrong??

b) KEf = umg * d?

2. Oct 29, 2008

### Rake-MC

Possibly the book is implying that when it gets to the bottom of the incline, it is on a horizontal surface, perhaps your final velocity is parallel to the incline?

b) looks ok to me, assuming KEf is the kinetic energy at the base of the incline

3. Oct 29, 2008

### Quincy

Suppose a roller coaster 45 m high has a speed of 1.7 m/s. If the average force of friction is equal to one-fifth of its weight, with what speed will it reach the the ground?

Relevant Equations: -Wnc = (KEf - KEo) + (PEf - PEo)

-- PEf cancels out

The equation becomes: (1/2)mvo2 + mgh - mg/5 = (1/2)mvf2

-- masses cancel out

= (1/2)(1.7)2 + (9.8)(45) - (9.8/5) = (1/2)vf2

Vf = 29.7 m/s

-- According to the book, the speed is 23 m/s... What am I doing wrong in this problem?

4. Oct 29, 2008

### Rake-MC

Just skimming over it, it looks like there's an error with your units in there:

Using that equation (providing the masses hadn't been cancelled {although it was fine to cancel them}) your units are: J + J - N = J
(correct me if I've interpreted it wrong).

5. Oct 29, 2008

### Quincy

Yes, the units are wrong, it should be (1/2)mvo2 + mgh - (mg/5 * d) = (1/2)mvf2

- the masses would still cancel out, and the equation would become:

(1/2)(1.7)2 + (9.8)(45) - (9.8/5 * 45) = (1/2)vf2

vf = 26.6 m/s... still not the right answer...

6. Oct 30, 2008

{Bump}