Conservation of energy problem(s)

[Quincy]

Homework Statement

A ski starts from rest and slides down a 22^o 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.

Homework Equations

-W_nc = (KE_f - KE_o) + (PE_f - PE_o)

The Attempt at a Solution

a)Since the hypotenuse of the incline is 75 m, and the angle relative to the horizontal is 22^o, the vertical height is 75 sin22^o = 28.1 m.
PE_f and KE_o cancel out, and the equation becomes:

PE_o = KE_f + W_nc

That simplifies to:

mgh = (1/2)mv² + umgcos22^o -- (u is coefficient of friction)

masses cancel out

(9.8)(28.1) = (1/2)v² + (0.09)(9.8)cos22^o

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

b) KE_f = umg * d?

 

[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 KE_f is the kinetic energy at the base of the incline

 

[Quincy]

Also, how about this:

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: -W_nc = (KE_f - KE_o) + (PE_f - PE_o)

-- PE_f cancels out

The equation becomes: (1/2)mv_o² + mgh - mg/5 = (1/2)mv_f²

-- masses cancel out

= (1/2)(1.7)² + (9.8)(45) - (9.8/5) = (1/2)v_f²

V_f = 29.7 m/s

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

 

[Rake-MC]

[...]
The equation becomes: (1/2)mv_o² + mgh - mg/5 = (1/2)mv_f²

[...]

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 canceled {although it was fine to cancel them}) your units are: J + J - N = J
(correct me if I've interpreted it wrong).

 

[Quincy]

Yes, the units are wrong, it should be (1/2)mv_o² + mgh - (mg/5 * d) = (1/2)mv_f²

- 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)v_f²

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

 

[Quincy]

{Bump}