Box and Spring on an Incline

  • Thread starter kliang1234
  • Start date
  • #1
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A 17.0 kg box slides 4.0 m down the frictionless ramp shown in the figure, then collides with a spring whose spring constant is 190 N/m.

Picture: http://session.masteringphysics.com/problemAsset/1000077659/2/knight_Figure_10_69.jpg

We also know:
a) What is the maximum compression of the spring? 2.36m I got this answer correct


My question is on b:

At what compression of the spring does the box have its maximum speed?

I was thinking this was a trick question. From the picture, the moment the box hits the spring, it slows down because the kinetic energy is transferred to spring potential energy. So i initially thought that the speed would be maximum right before it touches the spring. Therefore, compression of the spring would be at 0m.
That was incorrect.

Then I was thinking, maybe i was wrong and that maybe maximum velocity would be when acceleration is = 0.
To find the point when acceleration would be 0, i used:
(4m x 9.8m/s^2 x 17kg x sin(30)) = 333.2 N for work done while sliding down before hitting the spring

And we know F = kx

I was thinking that acceleration would equal 0 when kx = 333.2N because the compression would counteract the sliding down force.

With k = 190 i get x to get 1.75m
That was also incorrect

I have like 2 tries left on this stupid program.
Please someone help me out.
 

Answers and Replies

  • #2
Borek
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Before F from the string is higher than F from gravitation, box still accelerates, doesn't it?
 
  • #3
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if you meant spring, yes it still accelerates.
Thats why i thought to set kx = Fdsin30
but that was incorrect
 
  • #4
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I think I agree with your first guess. Also in that problem there is no instant in time at which acceleration is zero since the box would never remain at a constant velocity.
 
  • #5
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Acceleration would be gsin30 if I'm not mistaken.
There would be a time where acceleration will be equal to zero and that is when the retarding force of the spring pushes back onto the box until the box stops and the spring expands and the box slides back up the incline.

Both my tries seem to carry some truth in my opinion but they are wrong.

should i be doing .5kx^2 = fdsin30 instead?

kx = fdsin30 is for an instance where the spring force equals force of box sliding down the incline.

.5kx^2 would make more sense because the spring force isnt constant, there should be a summation of it so .5x^2 = fdsin30 makes sense too but im really scared to try that answer without some guidance
 
  • #6
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Yeah, acceleration would be gsin30 when the box is travelling down the slope. The acceleration the box experiences when the spring slows it to rest is not zero. So the max velocity of the box is before it hits the spring. There is a slight difference between speed and velocity though perhaps thats the issue. Since the box is travelling down is that the negative direction?
 
  • #7
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scratch that last statement... but yeah the velocity should be max when it first comes into contact with the spring im not sure how that isnt correct.
 
  • #8
Borek
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but yeah the velocity should be max when it first comes into contact with the spring im not sure how that isnt correct.
No. After the box comes in contact with the spring it still accelerates. At this moment it is acceleration that starts to get smaller, not the speed. When the box starts to slow down it means acceleration changed sign - that means there was a moment when it was zero. Speed was growing to that moment.
 
  • #9
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ok, so i tried .5kx^2 = fdsin30

that wasn't it either. i'm down to my last try on masteringphysics :cry:
 
  • #10
13
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nevermind.. after thinking about it some more, i finally got it
 
  • #11
10
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Im curious what did you find the answer to be?
 

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