# Homework Help: Spring and box problem

1. May 6, 2010

### xstetsonx

A 8 kg box slides d = 4.4 m down the frictionless ramp shown in Figure CP10.69, then collides with a spring whose spring constant is 250 N/m. At what compression of the spring does the box have its maximum velocity?

i know that .5kx^2+.5mv^2=mg(d+x)sin theta
-------------spring--kinetic-----potential----

but what is next?

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2. May 6, 2010

### tiny-tim

Hi xstetsonx!

(have a theta: θ and try using the X2 tag just above the Reply box )
That's fine … now put v on the LHS, and maximise the RHS.

3. May 6, 2010

### xstetsonx

i am sorry LHS (left hand side)? if so v is already on the left hand side and how i max the right?

4. May 6, 2010

### tiny-tim

I meant on its own.

5. May 6, 2010

### xstetsonx

but how do i know what x is?

6. May 6, 2010

### tiny-tim

uhh? x is distance … you have a function of x, and you need to maximise that function

7. May 6, 2010

### xstetsonx

how do i do that?
do i take the derivative????

Last edited: May 6, 2010
8. May 6, 2010

### tiny-tim

That's right … it's a maximum (or minimum or inflection point) when the derivative is zero.

9. May 6, 2010

### xstetsonx

OMG that IS why my solution manual did the derivative....... this is back to calculus isn't it don't you just hate they tell you the solution without saying why

o btw do i do it in respect of x or v?

10. May 6, 2010

### tiny-tim

well, duh … it's a spring question …

so they spring it on you!
v is a function of x, and you want to maximise v,

so you differentiate v wrt x

(if you draw a graph, with v going up and x going across, then dv/dx is the slope, so the maximum for v will be when the slope is horizontal, ie zero)

11. May 6, 2010

### xstetsonx

god this problem sucks i still haven't been able to solve my equation and get the right answer-----(15.68)

12. May 6, 2010

### tiny-tim

hmm … i make it .01568

what equation did you get?

13. May 6, 2010

### xstetsonx

oops it is in cm and yours i think is in m. so ur answer is right. but do u mind show me how you get there?

14. May 6, 2010

### tiny-tim

15. May 6, 2010

### xstetsonx

i didn't get it it comes from the solution manual .....

16. May 6, 2010

### tiny-tim

Yes, I know that, but you said you tried and got the wrong answer …
… so show us what you did.

17. May 6, 2010

### xstetsonx

v=((-125x^2+39.2x+172.48)/(4))^.5

dv/dx=.5(-31.25x^2+9.8x+43.12)^-.5(62.5x+9.8)

definitely don't think that is right

Last edited: May 6, 2010
18. May 7, 2010

### tiny-tim

Hi xstetsonx!

(just got up :zzz: …)

what's wrong with that? (apart from the missing minus sign)

you now have dv/dx = 0 when x = 9.8/62.5

(also, maximising v is the same as maximising v2, so you could simply have gone for dv2/dx = 0 instead )

btw I should have mentioned before … if you don't like differentiating, you can just complete the square instead

19. May 7, 2010

### xstetsonx

sorry didn't have time to reply till now

the x^2 hates me it didn't work..... or i don't know how to use it

do you set that to 0???? and solve for x? and then.....wat?

what's wrong with that? (apart from the missing minus sign)

you now have dv/dx = 0 when x = 9.8/62.5

(also, maximising v is the same as maximising v2, so you could simply have gone for dv2/dx = 0 instead )

complete the square in what step?

btw I should have mentioned before … if you don't like differentiating, you can just complete the square instead [/QUOTE]

guess i don't know how to quote either......