Spring launches ball, skier friction on incline

[jrd007]

Conservation of Mechanical Energy
(1) A vertical spring (ignore it's mass), whose spring stiffness constant is 950 N/m, is attached to a table and is compressed down 0.150m. (a) What upaward speed can it give to a 0.30kg ball when released? (b) How high above it's orginal position (Spring compressed) will the ball fly?
(a) Okay this is what I did for part a, and my answer was close but not close enough...
1/2kx^2 = 1/2mv^2
.5*950N/m*.150m^2 = .5*.30kg*v^2
v = 8.4 m/s (but the book gave the answer of 8.3 m/s) did I do anything incorrect?
(b) 1/2Kx^2=mgh
.5*950*.150^2 = .3*9.8*h
h = 3.64 m <~~~ I got that one correct.
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.
I need some to help me with problem 2. I have drew free bodies already... can't figure out how to attempt it.

 

[arildno]

Conservation of Mechanical Energy
(1) A vertical spring (ignore it's mass), whose spring stiffness constant is 950 N/m, is attached to a table and is compressed down 0.150m. (a) What upaward speed can it give to a 0.30kg ball when released? (b) How high above it's orginal position (Spring compressed) will the ball fly?
(a) Okay this is what I did for part a, and my answer was close but not close enough...
1/2kx^2 = 1/2mv^2
.5*950N/m*.150m^2 = .5*.30kg*v^2
v = 8.4 m/s (but the book gave the answer of 8.3 m/s) did I do anything incorrect?

Yes.
You forgot to include a very important type of potential energy in your equation for energy conservation here..

As for 2, what work must be done against the friction?

 

[jrd007]

A Very important type of PE? Huh? Thereis a missing variable?

 

[arildno]

Well, you haven't included the change in GRAVITATIONAL potential energy from when the spring is compressed and when it is not.

 

[jrd007]

So maybe...

mgh + 1/2kx^2 = 1/2mv^2
(.30kg*9.8*.150m) + (.5*950N/m*.150m^2) = .5*.30kg*v^2
11.1285 = .15 v^2
sqr of 74.19 = 8.61 <~~~ that isn't right...

 

[jrd007]

I think I may have got it... I believe I added the PE incorrectly. The equation should read:

1/2kx^2 = 1/2mv^2 + mgy

(.5*950N/m*.150m^2) = .5*.30kg*v^2 + (.30kg*9.8*.150m)
10.2465 = .15v^2
sqr of 68.31 = 8.26 = 8.3 m/s

so, that is the correct way?

 

[jrd007]

(b) 1/2Kx^2=mgh
.5*950*.150^2 = .3*9.8*h
h = 3.64 m

that is correct also?