Speed of skiis (someone check my work, please)

[jrd007]

Law of Conservation of Energy
(2) A ski starts from rest and slides down a 22 degree incline 75 m long.
(a) If the coeffiecent of friction is .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 coeffiecent of friction, how far will the ski travel along the level? Use energy methods. correct answer: 2400 m
------------------
My thoughts were to use the equation Wnc = KE + PE

Seeing as there is no potential energy on the way down I now have Wnc = KE

Then I had to find the work acting on the skiis... so...

Work(weight) + Work(friction) = (1/2)mv^2

Weights force would be (sin22 x m x g)
frictions would be (.090 x N) = (.090 x cos22 mg)

I would say the m's cancel out. So then we substitute back into Work(weight) + Work(friction) = (1/2)v^2

((sin22 x g) x 75m x cos0) + ((.090 x cos22 x g) x 75m x cos180) = 1/2v^2

so I get...

275 + (-61) = .5V^2
214/.5 = v^2
sq root of 428 = v
v = 20.6 = 21.0 Answer in Book: 21 m/s

Part B

Would I use a similar equation but instead of the Wnc = KE it would be Worknc = PE
Work nc = mgh ?

 

[jrd007]

can someone please help me? my exam is friday... :(

 

[lostdaytomorrow]

Can't you just plug in the 21 m/s for an initial velocity somewhere?

 

[jrd007]

Maybe I could use kinematics?

v2 = vo2 +2a(x-xo) ?

That doesn't work...

I was thinking since it says at the bottom of the incline there would be no potential engery so I could use

PE = KE but that does not work either.

I cannot figure out an equation in which I am trying to find distance

 

[jrd007]

also tried first finding a so I could use kinematics.

a = V^2/75
21^2/75 = 5.88

Then I used:

v2 = vo2 +2a(x-xo)
0 = 21^2 +2(5.88)(x-75m)
-441 = 11.76(x-75)
-36 = x-75
39... and the correct answer is 2400, what haven't I tried?

 

[Doc Al]

Several ways to attack part B:

Start by finding the friction force.

Energy methods: Work done by friction = change in Energy

Kinematics: Use the force to find the acceleration.

 

[jrd007]

Finding Friction force would be, as previous stated, (but I do not have a mass, so how can I find the friction force?

How do I find the friction force though?

I do not have a mass...

Friction force = .090 x ?

 

[jrd007]

would it be friction force = .090 x 9.8 m/s = .882 ?

My problem is writing the questions out...

(.882 N)(75 m) cos0 = 61.65 J

61.65 J = change in energy? What does that mean?

Sorry, Doc Al. I am just getting lost here.

 

[Doc Al]

Finding Friction force would be, as previous stated, (but I do not have a mass, so how can I find the friction force?

Just find it in terms of the unknown mass. You'll find that you don't need to know the mass to solve the problem. (Sorry if I wasn't clear before.)

How do I find the friction force though?

I do not have a mass...

Friction force = .090 x ?

Friction will equal [itex]\mu N[/itex]. What's N?

 

[jrd007]

N = W = mg.

f = [itex]\mu mg[/itex]

So would it just be [itex]\mu g[/itex]

 

[Doc Al]

N = W = mg.
f = [itex]\mu mg[/itex]

Exactly right.

So would it just be [itex]\mu g[/itex]

Not sure why you dropped the m.

In any case, now you are ready to solve the problem using energy methods or using kinematics. Do it both ways!

Energy method: You have the friction force and the distance, so what's the work done by friction? (Don't forget that it's negative.) And what's the change in KE? ([itex]{KE}_f - {KE}_i[/itex])

Kinematics: You have the friction force, so what's the acceleration?

 

[jrd007]

Work by f = change

([itex]\mu mg[/itex])(d)cos(180) = 1/2mv^2? -1/2mv^2

([itex]\mu mg[/itex])(d)cos(180) = -220 - 220 (440)

That still does not work... am I solving for the D?

 

[jrd007]

With Kinematics: F = ma
(.09 m g) = m (a) ---- m's cancel out
.09 x 9.8 = a
.882 m/s

v2 = vo2 +2a(x-xo)
21^2 = 0 +2(.882) (x-xo)
441/1.764 = (x -xo)
250...

not getting the answer either way. :(

 

[Doc Al]

([itex]\mu mg[/itex])(d)cos(180) = 1/2mv^2? -1/2mv^2

This is correct. What's the final KE? You found (in part a) the initial speed the skier has entering the level section. I'd rewrite it like this:
[tex]- \mu m g D = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2[/tex]

 

[Doc Al]

With Kinematics: F = ma
(.09 m g) = m (a) ---- m's cancel out
.09 x 9.8 = a
.882 m/s

v2 = vo2 +2a(x-xo)
21^2 = 0 +2(.882) (x-xo)
441/1.764 = (x -xo)
250...
not getting the answer either way. :(

This is correct. (Guess what? The book's answer is wrong!) Now do it using energy methods and verify that you get the same answer: D = 250 m.

(I didn't check the answer to part a; I assume that one's correct.)

 

[jrd007]

The following code was used to generate this LaTeX image:



[tex]- \mu m g D = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2[/tex]


So I substitute with numbers

-.09 x 9.8 x D = (1/2)(0 cause it will be when skii stops) - (1/2)(21^2)


-.882 x D = -220.5
D = 250 m

And you're saying the book is wrong?

 

[Doc Al]

And you're saying the book is wrong?

That's what I'm saying.