Final velocity of a block on a spring pulled downhill

  • #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.
 
Last edited:

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
15,267
4,244
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
Elaren
6
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 ## !​

##\ ##
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.
 
Last edited:
  • #4
erobz
Gold Member
1,478
659
I can’t figure out how the two (diagram and problem statement) relate to one another.
 
  • #5
Elaren
6
2
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
BvU
Science Advisor
Homework Helper
15,267
4,244
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
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
12,575
5,722
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
Elaren
6
2
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
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
12,575
5,722
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
BvU
Science Advisor
Homework Helper
15,267
4,244
with a constant force
$$f(t)=F_0\cos\omega t \qquad ? $$

##\ ##
 
  • #11
Elaren
6
2
$$f(t)=F_0\cos\omega t \qquad ? $$

##\ ##
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.
 
  • #12
Elaren
6
2
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.
 

Suggested for: Final velocity of a block on a spring pulled downhill

  • Last Post
Replies
6
Views
457
Replies
8
Views
355
Replies
10
Views
618
  • Last Post
Replies
27
Views
3K
  • Last Post
Replies
6
Views
425
  • Last Post
Replies
13
Views
431
Replies
11
Views
408
  • Last Post
Replies
22
Views
796
  • Last Post
Replies
2
Views
132
Top