Spring and box problem

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In summary, the conversation discusses a problem involving a box sliding down a ramp and colliding with a spring, in which the compression of the spring at maximum velocity needs to be determined. The formula .5kx^2+.5mv^2=mg(d+x)sin theta is provided, and the next step is to put v on the left hand side and maximize the right hand side. This is done by taking the derivative and setting it to zero. The discussion also touches on completing the square as an alternative method for solving the problem.
  • #1
xstetsonx
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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
Hi xstetsonx! :smile:

(have a theta: θ and try using the X2 tag just above the Reply box :wink:)
xstetsonx said:
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. :smile:
 
  • #3
i am sorry LHS (left hand side)? if so v is already on the left hand side and how i max the right?
 
  • #4
I meant on its own.
 
  • #5
but how do i know what x is?
 
  • #6
xstetsonx said:
but how do i know what x is?

uhh? :confused: x is distance … you have a function of x, and you need to maximise that function
 
  • #7
how do i do that?
do i take the derivative?
 
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  • #8
That's right … it's a maximum (or minimum or inflection point) when the derivative is zero. :smile:
 
  • #9
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?
 
  • #10
xstetsonx said:
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! :biggrin:
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 :smile:

(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
god this problem sucks i still haven't been able to solve my equation and get the right answer-----(15.68)
 
  • #12
hmm … i make it .01568 :confused:

what equation did you get?
 
  • #13
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
erm :redface:you show us how you got your answer
 
  • #15
i didn't get it it comes from the solution manual ...
 
  • #16
xstetsonx said:
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 …
xstetsonx said:
i still haven't been able to solve my equation and get the right answer-----(15.68)
… so show us what you did. :smile:
 
  • #17
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
 
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  • #18
Hi xstetsonx! :smile:

(just got up :zzz: …)
xstetsonx said:
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 :wink:)

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

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

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

btw I should have mentioned before :redface: … if you don't like differentiating, you can just complete the square instead :wink:
 
  • #19
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 :wink:)
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 :smile:

(also, maximising v is the same as maximising v2, so you could simply have gone for dv2/dx = 0 instead :biggrin:)
complete the square in what step?

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

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

1. What is the "spring and box problem"?

The "spring and box problem" is a classic physics problem that involves a horizontally placed spring attached to a box. The box is pulled to one side and then released, causing it to oscillate back and forth due to the spring's restoring force.

2. What are the key factors that affect the motion of the box in the "spring and box problem"?

The key factors that affect the motion of the box in the "spring and box problem" are the mass of the box, the spring constant of the spring, and the amplitude of the initial displacement of the box.

3. How does the spring constant affect the motion of the box in the "spring and box problem"?

The spring constant, represented by the letter k, determines how stiff or soft the spring is. A higher spring constant means a stiffer spring, which results in a faster oscillation of the box. On the other hand, a lower spring constant results in a slower oscillation.

4. What is the equation that describes the motion of the box in the "spring and box problem"?

The equation that describes the motion of the box in the "spring and box problem" is x = A cos(ωt), where x is the displacement of the box, A is the amplitude of the initial displacement, ω is the angular frequency, and t is time.

5. How does the "spring and box problem" relate to real-world applications?

The "spring and box problem" is a simplified model that can be applied to real-world situations, such as the motion of a car's suspension system or the oscillation of a mass on a spring. It also has applications in fields like earthquake engineering and seismology, where the motion of an object attached to a spring can be used to measure ground vibrations.

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