Final velocity of a block on a spring pulled downhill

In summary: The problem is missing a fundamental equation or relationship between the variables. In summary, the spring stretches as it goes down the slope.
  • #1
Elaren
6
2
Homework Statement
Calculate the final speed of the block, if the force has a magnitude of 200N and the displacement is 2m along the x-direction. The system starts from rest.
Relevant Equations
##W_{nc}=\Delta KE + \Delta PE##
phpCT1b1B.png

Note: wording is ambiguous so I assumed spring started from equilibrium, in which case it stretches as we go downslope. Final height (at lower point on slope) is 0.

Distance along slope = Distance the spring stretches = d= ##s_f## = ##2/cos{\theta}## =2.13
Height change = h = ##2 tan{\theta}## = .7

##Fd=\frac{1}{2} m v_f^2 - mgy_i + \frac{1}{2} k s_f^2##
##426=50 v_f^2 - 686 + 9074##
Leads to a square root of a negative number, so no real solution
I think this problem is wrong as written with a spring that strong. I found this same diagram being used for a perfectly reasonable oscillation problem, so I think it was taken and reused incorrectly.
 
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  • #2
Hi,

Are you assuming we pick up the preceding part of this exercise by means of telepathy ? Please post the complete problem statement!

Check out the PF guidelines

And

:welcome: ##\qquad ## !​

##\ ##
 
  • #3
BvU said:
Hi,

Are you assuming we pick up the preceding part of this exercise by means of telepathy ? Please post the complete problem statement!

Check out the PF guidelines

And

:welcome: ##\qquad ## !​

##\ ##
That is the homework complete statement. There is no earlier part. I included the picture in the main text. No way I can see to add an image to the problem statement.
 
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  • #4
I can’t figure out how the two (diagram and problem statement) relate to one another.
 
  • #5
erobz said:
I can’t figure out how the two (diagram and problem statement) relate to one another.
Yes I agree the problem is not well written. I took it to be the block moving down slope, with the 2 m being along the x axis (not the slope), with the 200N pointed downslope, the spring pulling up slope.
 
  • #6
Elaren said:
That is the homework complete statement. There is no earlier part. I included the picture in the main text. No way I can see to add an image to the problem statement.

Sorry, I can't make sense of this problem statement.
Which probably means I have the same problem you have: what's going on ?
##mg\sin\theta## plus the 400 N ##F_0## is way too small to ever stretch the spring by 2 m. Let alone that ##\omega## is completely unknown it could well be enormous.

In addition the ##x## direction is unassigned -- leaving you to guess whether it's along the incline or along the horizontal.

Someone is wasting your time !

##\ ##
 
  • #7
You can add images using the "Attach files" link, lower left. The statement of the problem requires a certain level of mind reading that is beyond me. I have a few questions/observations.

1. What makes the speed "final"? It looks like the mass will oscillate up and down the incline, so the speed varies with position. A reasonable reformulation of the question would be to find the speed of the block when its displacement is 2 m along the x-direction.
2. Yes, but which direction is "x"? Is it the horizontal or the down incline direction? The figure lacks axes. OP assumes the former.
3. There is force ##f(t)=F_0\cos\omega t## shown acting on the block. It appears that we have a driven harmonic oscillator with ##F_0=200~##N.

I am as befuddled as @BvU who posted as I was finishing this.
 
  • #8
I put in the image of the problem in the only spot I could figure out how to do so. I am confused as to why people seem to be having trouble with the post itself.

As to the problem being confusing, I agree. I mentioned the assumptions I was making in my work section (where the axis are, etc). This image was definitely cribbed from an oscillation problem and modified to be used as an energy problem (badly) with a constant force and asking the final speed after the first 2m movement from rest.
 
  • #9
Elaren said:
I put in the image of the problem in the only spot I could figure out how to do so. I am confused as to why people seem to be having trouble with the post itself.

As to the problem being confusing, I agree. I mentioned the assumptions I was making in my work section (where the axis are, etc). This image was definitely cribbed from an oscillation problem and modified to be used as an energy problem (badly) with a constant force and asking the final speed after the first 2m movement from rest.
I am having problems with the numbers, not necessarily with the post.
The down-incline component of the weight is
##mg \sin20^o=335~##N.
The maximum value of the driving force is 200 N. The two added together give a down-incline maximum force of 535 N. When the spring is stretched by 2 m, the up-incline force is ##kx=8000~##N which is much larger than the relatively pitiful down-incline force. This means that the spring will never be stretched by 2 m so what are we supposed to find?

Something is amiss.
 
  • #10
Elaren said:
with a constant force
$$f(t)=F_0\cos\omega t \qquad ? $$

##\ ##
 
  • #11
BvU said:
$$f(t)=F_0\cos\omega t \qquad ? $$

##\ ##
The problem says the force is 200N, and variable forces are not part of this course, so the marking on the picture is ignored I guess.
Yes this is the amount of care I have come to expect in figure reuse from this instructor.
 
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  • #12
kuruman said:
I am having problems with the numbers, not necessarily with the post.
The down-incline component of the weight is
##mg \sin20^o=335~##N.
The maximum value of the driving force is 200 N. The two added together give a down-incline maximum force of 535 N. When the spring is stretched by 2 m, the up-incline force is ##kx=8000~##N which is much larger than the relatively pitiful down-incline force. This means that the spring will never be stretched by 2 m so what are we supposed to find?

Something is amiss.
This is why I said in my opening post that I thought the problem was invalid as written. Not the first time with this person.
I am mostly checking here to see if I am missing something.
 
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Related to Final velocity of a block on a spring pulled downhill

1. What is the final velocity of a block on a spring pulled downhill?

The final velocity of a block on a spring pulled downhill can be calculated using the equation v = √(2kx/m), where k is the spring constant, x is the displacement of the block, and m is the mass of the block.

2. How does the spring constant affect the final velocity of the block?

The spring constant directly affects the final velocity of the block. A higher spring constant means that the spring is stiffer and will produce a larger force, resulting in a higher final velocity for the block.

3. Does the mass of the block have an impact on the final velocity?

Yes, the mass of the block does have an impact on the final velocity. A heavier block will require more force to accelerate, resulting in a lower final velocity compared to a lighter block pulled by the same spring.

4. How does the displacement of the block affect the final velocity?

The displacement of the block also has an impact on the final velocity. The further the block is pulled down, the more potential energy it has, resulting in a higher final velocity when released by the spring.

5. Are there any other factors that can affect the final velocity of the block on a spring?

Other factors that can affect the final velocity of the block on a spring include air resistance and friction. These factors can slow down the block and decrease its final velocity. The surface on which the block is sliding can also affect its final velocity.

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