Energy to compress a spring 3 units

[Dustinsfl]

Consider a block on a frictionless surface. The force applied the block in the positive x direction is \(F = e^x\) and the force in the negative x is \(kx\) where k is the spring constant.

How is the energy required determined to move the block 3 units?

I set up Newton's 2nd but I don't think that is going to help:
\[
\ddot{x} + \frac{k}{m}x = \frac{1}{m}e^x
\]
Then I set up CoE assuming at 3 units, we only have \(PE = \frac{1}{2}kx^2\) and the block at rest only has \(KE = e^x\).
\[
e^x = \frac{1}{2}kx^2
\]
but this will only tell me the spring constant.

 

[topsquark]

Consider a block on a frictionless surface. The force applied the block in the positive x direction is \(F = e^x\) and the force in the negative x is \(kx\) where k is the spring constant.

How is the energy required determined to move the block 3 units?

I set up Newton's 2nd but I don't think that is going to help:
\[
\ddot{x} + \frac{k}{m}x = \frac{1}{m}e^x
\]
Then I set up CoE assuming at 3 units, we only have \(PE = \frac{1}{2}kx^2\) and the block at rest only has \(KE = e^x\).
\[
e^x = \frac{1}{2}kx^2
\]
but this will only tell me the spring constant.

For starters we have no given information to tell us that the block is not moving initially, or at least you didn't post it, so your analysis of the balancing point x may be in error. Anyway unless you need an equation of motion we can do this more directly by considering the work done on the box:
\(\displaystyle W = \int _0 ^3 F \cdot dx\)

where \(\displaystyle F = e^x - kx\). Be careful of the direction of F...it will be positive on some interval and negative on another. I leave that one to you.

-Dan

 

[Dustinsfl]

For starters we have no given information to tell us that the block is not moving initially, or at least you didn't post it, so your analysis of the balancing point x may be in error. Anyway unless you need an equation of motion we can do this more directly by considering the work done on the box:
\(\displaystyle W = \int _0 ^3 F \cdot dx\)

where \(\displaystyle F = e^x - kx\). Be careful of the direction of F...it will be positive on some interval and negative on another. I leave that one to you.

-Dan

Can you explain this more?

 

[Ackbach]

topsquark's approach is probably the simplest. We know that
$$W=\int \mathbf{F} \cdot d\mathbf{r},$$
technically. So if the net force is in the direction of travel, we'll get positive work done by that force, otherwise negative. In this case, the spring force is always to the left, and the exponential force is always to the right. The effect of the spring force is simply going to be to increase the amount of work required. So you must compute
$$W=\int_0^3 (e^{x}-kx) \, dx.$$
Does that answer your question?

 

[Dustinsfl]

topsquark's approach is probably the simplest. We know that
$$W=\int \mathbf{F} \cdot d\mathbf{r},$$
technically. So if the net force is in the direction of travel, we'll get positive work done by that force, otherwise negative. In this case, the spring force is always to the left, and the exponential force is always to the right. The effect of the spring force is simply going to be to increase the amount of work required. So you must compute
$$W=\int_0^3 (e^{x}-kx) \, dx.$$
Does that answer your question?

I was mainly asking about the positive negative switch. That is determined by k though which we don't know.

 

[Ackbach]

Right. If $k<e$, you get no switching, if $k=e$ you get one intersection, and if $k>e$ you get two switches. However, I don't think the force in the work integral is a magnitude. That is, I don't think you have to split the integral up into regions, and use different expressions depending on where the two graphs intersect. Work can definitely be negative. In other words, you can compute
$$W=\int_0^3 \left( e^{x}-kx \right) \, dx,\quad \text{not} \quad \int_0^3 \left| e^x - kx \right| \, dx.$$

 

[Dustinsfl]

Right. If $k<e$, you get no switching, if $k=e$ you get one intersection, and if $k>e$ you get two switches. However, I don't think the force in the work integral is a magnitude. That is, I don't think you have to split the integral up into regions, and use different expressions depending on where the two graphs intersect. Work can definitely be negative. In other words, you can compute
$$W=\int_0^3 \left( e^{x}-kx \right) \, dx,\quad \text{not} \quad \int_0^3 \left| e^x - kx \right| \, dx.$$

Also, how would I go from work to energy afterwards?

 

[Ackbach]

Also, how would I go from work to energy afterwards?

The work computed in this way would BE the energy: work has units of joules, the same as energy.