Conservation of energy: Question about simplifying this equation (cart and spring)

  • Thread starter adams_695
  • Start date
  • #1
adams_695
16
1
Homework Statement:
Unable to solve for solution.
Relevant Equations:
Kinematic energy and spring energy
Am unable to understand how in the final answer the X2^2 gets canceled out without being inside the radical. It doesn’t make sense to me.

Any help explaining would be much appreciated as I am stuck.

Also am happy to provide more information if needed.
1F3542BE-362C-461B-A0DC-AFD5FD619DFC.png
 
Last edited by a moderator:

Answers and Replies

  • #2
adams_695
16
1
2A555E61-7BE8-47EB-B8E2-58154607E046.png

Full visibility
 
  • #3
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
12,641
5,782
Can you solve the equation$$\frac{1}{2}kx_2^2=\frac{1}{2}mv_1^2$$for ##v_1## in terms of the other quantities? What do you get?
 
  • #4
adams_695
16
1
Can you solve the equation$$\frac{1}{2}kx_2^2=\frac{1}{2}mv_1^2$$for ##v_1## in terms of the other quantities? What do you get?

Yes it’s just x2 to take out the 1/2 after factoring on both sides then divide by M.

I’m more trying to understand why the x is left outside and isn’t inside the radical also. In this one here it’s left outside when usually it’s left inside.

Is there a particular reason?
 
  • #5
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
12,641
5,782
The only reason I can think of is "simplify as much as you can to enhance readability and ease of calculation". Remember that $$\omega=\sqrt{\frac{k}{m}}$$ so that$$v_1=\sqrt{\frac{k}{m}}x_2=\omega \ x_2$$ The last result is simpler and more readable. Besides, if you want to find a number for ##v_1## and you put ##x_2^2## under the radical, you will have to square ##x_2## and then take its square root.
 
  • #6
jbriggs444
Science Advisor
Homework Helper
11,586
6,242
The only reason I can think of is "simplify as much as you can to enhance readability and ease of calculation". Remember that $$\omega=\sqrt{\frac{k}{m}}$$ so that$$v_1=\sqrt{\frac{k}{m}}x_2=\omega \ x_2$$ The last result is simpler and more readable. Besides, if you want to find a number for ##v_1## and you put ##x_2^2## under the radical, you will have to square ##x_2## and then take its square root.
Note that ##v^2=x^2##, simplifies to ##v=\pm x##. Be careful not to simplify one of the two solutions out of existence.
 
  • #7
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
12,641
5,782
Note that ##v^2=x^2##, simplifies to ##v=\pm x##. Be careful not to simplify one of the two solutions out of existence.
When a numerical answer is required, there should be enough information to select the solution that makes physical sense. The figure shows the cart moving to the right before hitting the spring therefore it is safe to conclude that the velocity is positive. That point should have been clarified in the solution.

I am bothered more by the solution's last pronouncement, "Elastic potential energy is always measured from the unstretched or uncompressed length of the spring." It unnecessarily complicates the analysis of the vertical mass-spring system and undermines the idea that the choice of the reference point of potential energy is arbitrary.
 

Suggested for: Conservation of energy: Question about simplifying this equation (cart and spring)

Replies
7
Views
505
Replies
15
Views
99
Replies
6
Views
338
  • Last Post
Replies
6
Views
440
Replies
20
Views
810
Replies
20
Views
361
  • Last Post
Replies
13
Views
451
  • Last Post
Replies
4
Views
393
Replies
21
Views
420
Top