How Does a Canister's Speed Change Along a Frictionless Incline?

[Gold3nlily]

Homework Statement

In Fig. 8-44, a spring with k = 170 N/m is at the top of a frictionless incline of angle θ = 37.0°. The lower end of the incline is distance D = 1.00 m from the end of the spring, which is at its relaxed length. A 2.00 kg canister is pushed against the spring until the spring is compressed 0.200 m and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

Image:

http://edugen.wiley.com/edugen/courses/crs4957/halliday9118/halliday9118c08/image_n/nt0051-y.gif

The answer from the book is:
(a) 2.40 m/s; (b) 4.19 m/s

Homework Equations

Emec1 = Emec2
Us = 1/2 k x^2
kE =1/2 m v^2
Ug = mgh
h = sin@*D

@=37
k=170 N/m
D=1m
m=2kg
d=0.2m

The Attempt at a Solution

(a) What is the speed of the canister at the instant the spring
returns to its relaxed length ?

My answer is close to the correct answer at 2.9m/s but I
think it should be closer...

Emec1 = Emec2
Ke + U = KE + U
0 + 1/2 kd^2 = 1/2mv^2 +mgh
h = sin@*D
1/2 kd^2 = 1/2mv^2 +mg sin (40)*D
mv^2= 1/2 k(0.2)^2 - (2)(9.8)sin (40)*1
v= sq(-16.79/2)
I also think I did something wrong b/c had to eliminate negative sign...
v= 2.9m/s

What could I have done differently?

(b) What is the speed of the canister when it reaches the lower end of the incline?
u bottom = 0; b/c h = 0
Emec1 = Emec2
Ke + U = KEb + Ub
1/2mv^2 +mg sin (40)*D = 1/2mv^2 + 0
cancel "m" and multiply by 2
v^2 + 2g sin (40)*1 = v^2
2.4^2 + 12.5986 = v^2
v = sq(18.358)
v = 4.28 m/s

This is still very close to the right answer but slightly different. (I Used
the books value for velocity to calculate part B) Sigh, I wish I was good
at this...

 

[tiny-tim]

Hi Gold3nlily!

(have a theta: θ and try using the X² icon just above the Reply box )

0 + 1/2 kd^2 = 1/2mv^2 +mgh

Hint: if you increase h, will that increase v, or decrease it?

 

[Gold3nlily]

Hi Gold3nlily!

Hint: if you increase h, will that increase v, or decrease it?

Hello Tim. Thank you for answering my question!

Thank you also for the tips to put [tex]\theta[/tex] and x².

Okay, so do you mean that it would be
0 + 1/2 kd² = 1/2mv² - mgh??

This way when h gets bigger, more of the spring potential energy will have to go into KE to balance and then velocity would get bigger. Yes? That makes sense, but can I also think of h as being negative becasue the block is moving down? If I did that would it mean that I am supposed to set my reference level (y=0) at the top?

When I do it this way I still don't get the exact answer.
mv²= 1/2 k (0.2)² + (2)(9.8)sin (40)*1
v = sq(7.99932) (YAY! this is real result!)
v = 2.83 m/s
(answer should be 2.4 m/s. Its closer, but am I making any other mistakes? ...I am not rounding any intermediates or anything.)

Thank you again Tim!

 

[tiny-tim]

mv²= 1/2 k (0.2)² + (2)(9.8)sin (40)*1
v = sq(7.99932)

How exactly do you get 7.99932 ?

 

[Gold3nlily]

How exactly do you get 7.99932 ?

mv²= 1/2 k (0.2)² + (2)(9.8)sin (40)*1
v²= (3.4 + 12.5986)/2 = 15.99983/2 = 7.99932
(b/c the mass is 2 kg and i divided by that?)
v = 2.83 m/s

 

[tiny-tim]

Hi Gold3nlily!

(just got up :zzz: …)

mv²= 1/2 k (0.2)² + (2)(9.8)sin (40)*1

ah, those 2s are confusing you …

you've left out a 0.2 at the end ​

(oh, and is it 37° or 40° ? )