Greetings to all.

I posted this question in the wrong newsgroup earlier, so I hope that
someone here will be able to shed some light on my questions. For those of
you who have the book, Serway & Faughn, College Physics 5th Edition, I am
referring to question 5.39.

"A skier starts from rest at the top of a hill that is inclined at 10.5
degrees with the horizontal. The hillside is 200m long, and the coefficient
of friction between snow and skis is 0.0750. At the bottom of the hill, the
snow is level and the coefficient of friction is unchanged. How far does
the skier move along the horizontal portion of the snow before coming to
rest?"

Possible solution:

The Fn of skier is mgcos10.5 while on hill and Fn of skier is mg on
horizontal portion.

Using F * s = Wnc = (KEf + PEf) - (KEi + PEi) and defining the bottom of
hill as zero level for potential energy, I used F * s =
(0.0750)(mgcos10.5)(200) = (1/2)m(v)^2 + 0 - 0 - mgh (h = 200sin10.5).

I determined v = 19.02 m/s.

Using F * s = Wnc = (KEf + PEf) - (KEi + PEi), I used F * s =
(0.0750)(mg)(s) = 0 + 0 - (1/2)(m)(19.02)^2 + 0.
s = 246.1m.

The answer in the back of the book is s = 289m.

Did I do the problem wrong? If so, can I solve it this way?

Thanks,
Will

"Will" wrote in message
...
Greetings to all.

I posted this question in the wrong newsgroup earlier, so I hope that
someone here will be able to shed some light on my questions. For
those of
you who have the book, Serway & Faughn, College Physics 5th Edition, I
am
referring to question 5.39.

"A skier starts from rest at the top of a hill that is inclined at
10.5
degrees with the horizontal. The hillside is 200m long, and the
coefficient
of friction between snow and skis is 0.0750. At the bottom of the
hill, the
snow is level and the coefficient of friction is unchanged. How far
does
the skier move along the horizontal portion of the snow before coming
to
rest?"

Possible solution:

The Fn of skier is mgcos10.5 while on hill and Fn of skier is mg on
horizontal portion.

Using F * s = Wnc = (KEf + PEf) - (KEi + PEi) and defining the bottom
of
hill as zero level for potential energy, I used F * s =
(0.0750)(mgcos10.5)(200) = (1/2)m(v)^2 + 0 - 0 - mgh (h = 200sin10.5).

I determined v = 19.02 m/s.

Using F * s = Wnc = (KEf + PEf) - (KEi + PEi), I used F * s =
(0.0750)(mg)(s) = 0 + 0 - (1/2)(m)(19.02)^2 + 0.
s = 246.1m.

The answer in the back of the book is s = 289m.

PE minus friction energy loss on slope = friction energy loss on flat
mgh - umg*cos10.5*s1 = umg*s2
cancel mg
h - u*cos10.5*s1 = u*s2
200*sin10.5 - .075*cos10.5*200 = .075*s2
s2 = (36.45 - 14.75)/.075
s2 = 289 m

PH

On Sun, 29 Aug 2004, Will wrote:

Greetings to all.

I posted this question in the wrong newsgroup earlier, so I hope that
someone here will be able to shed some light on my questions. For those of
you who have the book, Serway & Faughn, College Physics 5th Edition, I am
referring to question 5.39.

"A skier starts from rest at the top of a hill that is inclined at 10.5
degrees with the horizontal. The hillside is 200m long, and the coefficient
of friction between snow and skis is 0.0750. At the bottom of the hill, the
snow is level and the coefficient of friction is unchanged. How far does
the skier move along the horizontal portion of the snow before coming to
rest?"

Possible solution:

The Fn of skier is mgcos10.5 while on hill and Fn of skier is mg on
horizontal portion.

Using F * s = Wnc = (KEf + PEf) - (KEi + PEi) and defining the bottom of
hill as zero level for potential energy, I used F * s =
(0.0750)(mgcos10.5)(200) = (1/2)m(v)^2 + 0 - 0 - mgh (h = 200sin10.5).

Excepting the error in sign (you have Wnc positive, but (KEf + PEf) is
less than (KEi + PEi)), yes (and it doesn't affect your answer)

I determined v = 19.02 m/s.

With g = 9.8 m/s^2, I get 20.622 m/s. You either used a different value of
g, or rounded some intermediate results (which is quite OK).

Using F * s = Wnc = (KEf + PEf) - (KEi + PEi), I used F * s =
(0.0750)(mg)(s) = 0 + 0 - (1/2)(m)(19.02)^2 + 0.
s = 246.1m.

OK.

The answer in the back of the book is s = 289m.

v = 20.622 m/v gives 289m. The difference is just due to round-off errors.
(Or a different value of g).

Did I do the problem wrong? If so, can I solve it this way?

Only wrong if there's a calculation error (rather than rounding errors)
giving you v = 19.02 m/s.

Why not do it the short way?

You have E0 = KEi + PEi = PEi

E1 ( = KE1) = E0 - Wslope

E2 = 0 = E1 - Wflat, so Wflat = E1.

Minimise intermediate calculations, with fewer chances for error. Unless
you need the intermediate results.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html

Thanks for the reply Timo!

Why not do it the short way?

Well, I did it the way that came to mind first. I am trying to re-learn
this stuff and things are going slowly.

You have E0 = KEi + PEi = PEi

E1 ( = KE1) = E0 - Wslope

E2 = 0 = E1 - Wflat, so Wflat = E1.

Minimise intermediate calculations, with fewer chances for error. Unless
you need the intermediate results.

You are absolutely right on this point! Many times I set the problem up
correctly only to get an incorrect answer because I messed up in the math.
Back when I did this on a regular basis, I was very methodical and precise.
I guess I feel as if time is running out.

I am studying for the GRE and might possibly take the Physics GRE sometime
in the future. I have been out of undergrad for 6 years and have not done
anything physics -wise since I graduated.

Thanks again!

Will

"Will" wrote in
:

Greetings to all.

I posted this question in the wrong newsgroup earlier, so I hope that
someone here will be able to shed some light on my questions. For
those of you who have the book, Serway & Faughn, College Physics 5th
Edition, I am referring to question 5.39.

"A skier starts from rest at the top of a hill that is inclined at 10.5
degrees with the horizontal. The hillside is 200m long, and the
coefficient of friction between snow and skis is 0.0750. At the bottom
of the hill, the snow is level and the coefficient of friction is
unchanged. How far does the skier move along the horizontal portion of
the snow before coming to rest?"

Possible solution:

The Fn of skier is mgcos10.5 while on hill and Fn of skier is mg on
horizontal portion.

Using F * s = Wnc = (KEf + PEf) - (KEi + PEi) and defining the bottom
of hill as zero level for potential energy, I used F * s =
(0.0750)(mgcos10.5)(200) = (1/2)m(v)^2 + 0 - 0 - mgh (h = 200sin10.5).

You need a minus sign in front of the term on the left side of the
equation. F is in the opposite direction of s. With that your solution
works.

I determined v = 19.02 m/s.

Using F * s = Wnc = (KEf + PEf) - (KEi + PEi), I used F * s =
(0.0750)(mg)(s) = 0 + 0 - (1/2)(m)(19.02)^2 + 0.
s = 246.1m.

The answer in the back of the book is s = 289m.

Did I do the problem wrong? If so, can I solve it this way?

Thanks,
Will





--
Steve Gray

The obvious ... Im such a moron.

Thanks!

" You need a minus sign in front of the term on the left side of the
equation. F is in the opposite direction of s. With that your solution
works.


--
Steve Gray