Unable to Understand How X2^2 Gets Canceled Out

[adams_695]

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.

 

[adams_695]

Full visibility

 

[kuruman]

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?

 

[adams_695]

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?

 

[kuruman]

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.

 

[jbriggs444]

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.

 

[kuruman]

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.