Need help setting up the problem.

[qball1982]

Hello,
I was looking to get some help with setting up this problem so I may solve it. For some reason I just don't see what to do to work towards a solution.

Homework Statement

A 6kg collar is allowed to slide over a frictionless pole whose height above the ground obeys the parabolic equation y=8-(1/2)x^2. Attached to the collar is a k=30N/m spring. The spring is 1 m when unstretched and connected in a way that the spring will always start at the origin. If the collar started from rest at (0,8), how fast will it be traveling at x=2? Hint: Don't forget gravity!Any help to point me in the right direction would be fantastic. Thank you.

 

[tiny-tim]

welcome to pf!

hello qball1982! welcome to pf!

(try using the X² button just above the Reply box )

hint: try conservation of energy …

kinetic energy + gravitational potential energy + elastic potential energy = constant …

what do you get? ​

 

[qball1982]

Excellent! Thank you tiny-tim!

I worked this out using the formula: K₁+U₁=K₂+U₂
where U= mgy+1/2kx²

Crunching the numbers I get :
0+1205.4=K₂+778.05
427.35=K₂
427.35=1/2(6)v²

then finally: v=11.935m/s

Does this seem like the right path?

 

[tiny-tim]

hi qball1982!

(just got up :zzz:)

yes that's the correct method

(but we need to see your detailed calculations if you want us to check the actual figures … for example, did you use √(x² + y²) in the spring PE?)

 

[qball1982]

Hello tiny-tim!

I suppose exact calculations would help. Sorry about that.

Spring displacement calculations:
x₁ = 8-1 = 7 meters
x₂ = √(2²+6²) -1 = 5.32 meters
And of course the -1 in each is due to the springs unstretched length of 1 meter.

Spring U_el Calculations:
U_el1= 1/2(30N/m)(7m)² = 735N/m
U_el2= 1/2(30N/m)(5.32m)² = 425.25N/m

Gravitational Potential Energy Calculations:
U_g1= (6kg)(9.8m/s²)(8m)=470.4
U_g2= (6kg)(9.8m/s²)(6m)= 352.8

To solve:
K₁+U<sub>el1[/SUB+Ug1=K2+Uel2+Ug2

0+ 735+ 470.4 =K2+ 425.25 + 352.8
1205.2 = K2 + 778.05
427.15 =K2
427.15 = 1/2(m)(v²)
427.15 = 3(v²)
142.38 = (v²)
v=√(142.38)
v = 11.93 m/s

I think it looks right but I am also wrong quite a bit too.

Thanks again for all the help!</sub>

 

[tiny-tim]

yes that looks ok

just one thing, you could save time and make it look neater if instead of doing all this …

U_el1= 1/2(30N/m)(7m)² = 735N/m
U_el2= 1/2(30N/m)(5.32m)² = 425.25N/m

U_g1= (6kg)(9.8m/s²)(8m)=470.4
U_g2= (6kg)(9.8m/s²)(6m)= 352.8

you just work out U_el1 - U_el2 = 1/2(30N/m){(7m)² - (5.32m)²}

(and the same for U_g)