What Compression Level Gives Maximum Velocity to a Sliding Box on a Spring Ramp?

[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?

 

[tiny-tim]

Hi xstetsonx!

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

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

That's fine … now put v on the LHS, and maximise the RHS.

 

[xstetsonx]

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

 

[tiny-tim]

I meant on its own.

 

[xstetsonx]

but how do i know what x is?

 

[tiny-tim]

but how do i know what x is?

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

 

[xstetsonx]

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

 

[tiny-tim]

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

 

[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 whyo btw do i do it in respect of x or v?

 

[tiny-tim]

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

well, duh … it's a spring question …

so they spring it on you! ​

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

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)

 

[xstetsonx]

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

 

[tiny-tim]

hmm … i make it .01568 …

what equation did you get?

 

[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?

 

[tiny-tim]

erm … you show us how you got your answer

 

[xstetsonx]

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

 

[tiny-tim]

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

Yes, I know that, but you said you tried and got the wrong answer …

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

… so show us what you did.

 

[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

 

[tiny-tim]

Hi xstetsonx!

(just got up :zzz: …)

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

(please use the X² tag just above the Reply box )

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

 

[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

(please use the X² tag just above the Reply box )
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...