Am I underthinking this work problem?

[toesockshoe]

Homework Statement

A mass m is dropped from rest above a relaxed spring of stiffness k a distance D. Find the position from where it was released where the mass attains its maximum velocity and find that maximum velocity

Homework Equations

$$W_{net}=\Delta E$$

The Attempt at a Solution

ok... so obviously the mass is going to be at maximum velocity when it as distance D from the place of drop right? I stated that we can bascially ignore the spring in the problem because at the maximum velocty, the mass will JUST hit the spring and the spring won't have any impact or do any work on the mass.

so here is what i did;

SYSTEM: MASS and EARTH
$$W_{net}=\Delta E$$

There is no work in this problem so,

$$0=\Delta GPE + \Delta KE$$
$$0 = \frac{1}{2}mv_f^2-mgD$$
$$\frac{1}{2}mv_f^2 = mgD$$
$$v_f = \sqrt{2gD}$$
am i missing something crucial?

 

[toesockshoe]

ugh, also can someone tell me why everything I type in latex is centered? I use the same code for mathhelpforum and it works fine ther. here is the code I used for the last latex equation i wrote:

[ tex]v_f=\sqrt{2gD} [ /tex]whats wrong with that above code that that makes everything centered?

 

[insightful]

Consider a limiting case of small d and small k. That is, the weight has basically no time to accelerate before hitting a very weak spring.

 

[haruspex]

Given that the first thing it asks for is the position the mass was dropped from, I suggest D is not what one would assume. Perhaps it is the maximum distance the mass falls?
Either way, as Insightful says, you are wrong about the max speed being where it first contacts the spring. Think about acceleration.

 

[toesockshoe]

Given that the first thing it asks for is the position the mass was dropped from, I suggest D is not what one would assume. Perhaps it is the maximum distance the mass falls?

well the longer time it falls for the higher its velocity is correct? D is the distance from the position of release to the top of the spring (so just before the mass would contact the spring) so wouldn't that be the position where the velocity would be maximum?

 

[haruspex]

well the longer time it falls for the higher its velocity is correct? D is the distance from the position of release to the top of the spring (so just before the mass would contact the spring) so wouldn't that be the position where the velocity would be maximum?

No. See my edited reply.

 

[toesockshoe]

No. See my edited reply.

yeah, I am not understanding where I'm going wrong. the mass is in free fall... so the acceleration is 9.8. the acceleration is going to be 9.8 until it hits the spring where it's kinetic energy is transferred to elastic potential energy. do you mind telling me how the mass is not at its max velocity right before it hits the spring?

 

[haruspex]

yeah, I am not understanding where I'm going wrong. the mass is in free fall... so the acceleration is 9.8. the acceleration is going to be 9.8 until it hits the spring where it's kinetic energy is transferred to elastic potential energy. do you mind telling me how the mass is not at its max velocity right before it hits the spring?

Suppose it has hit the spring and depressed it a small distance, so the spring force is still quite small. Which way is it accelerating at this point?

 

[toesockshoe]

Suppose it has hit the spring and depressed it a small distance, so the spring force is still quite small. Which way is it accelerating at this point?

oh... ok. it is still accelerating downward. so my answer only work for springs that have a pretty large k value correct? (when the spring force is greater than mg).

 

[toesockshoe]

Suppose it has hit the spring and depressed it a small distance, so the spring force is still quite small. Which way is it accelerating at this point?

how do i find when the spring force is greater than the gravitational force?

through a side F=ma problem, found that:

$$k =\frac{m(a+g)}{x}$$

or

$$a = \frac{mg-kx}{m}$$

is that helpful in solving the priblem?

 

[haruspex]

how do i find when the spring force is greater than the gravitational force?

through a side F=ma problem, found that:

$$k =\frac{m(a+g)}{x}$$

or

$$a = \frac{mg-kx}{m}$$

is that helpful in solving the priblem?

Yes. By that, when is a zero?

 

[toesockshoe]

Consider a limiting case of small d and small k. That is, the weight has basically no time to accelerate before hitting a very weak spring.

alright. i fixed my answer a bit. doing an F=ma problem I did the follwing with the mass as the system and the y going down...

$$F_g - F_{el} = ma$$
we get $$a=\frac{mg-kx}{m}$$

we want to find the point at which a turns from positive to negative (becuase that's when acceleration changed directions)

so $$mg-kx > 0$$
is true as long as
$$x< \frac{mg}{k}$$ correct? where x is the length compression.

now we do the entire problem again:

now I think that the distance where velocity is greatest is the following:

$$D+\frac{mg}{k}$$

system: mass, earth, and spring:

work = change in energy
$$0 = \Delta GPE + \Delta EPE + \Delta KE$$
$$= -mg(D+\frac{mg}{k}) + \frac{m^2g^2}{2k}+ \frac{1}{2}mv^2$$
$$v =\sqrt{-2\bigg{(} \frac{-mg(D+\frac{mg}{k}) + \frac{m^2g^2}{2k}}{m}\bigg)}$$

 

[toesockshoe]

Yes. By that, when is a zero?

is my above solution correct?

 

[haruspex]

is my above solution correct?

You have some sign issues (is the change in GPE positive or negative?), and mg/2 is not an energy, it' a force.

 

[toesockshoe]

wait wait. don't correct me. i see an error.

 

[toesockshoe]

You have some sign issues (is the change in GPE positive or negative?), and mg/2 is not an energy, it' a force.

i think i fixed it now.

 

[insightful]

Check your final rearrangement to get v.

 

[toesockshoe]

Check your final rearrangement to get v.

my bad. there should be a negative inside the radical right?

 

[insightful]

Why is 1/2mv^2 under the radical?

 

[toesockshoe]

Why is 1/2mv^2 under the radical?

oh that's an accident as well. i forgot to take it out... yeah i make way too many stupid mistakes