F=ma useful or not?

[SaintJude]

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A 4- kg mass slides along a frictionless floor at 0.5 m/s until it collides with a spring. The mass comes to rest when the spring is compressed 5 cm. What is the force constant of the spring? Answer is 400 N/m.

Here's what I'm thinking.

To solve this problem, one ought NOT to use F=m.a=−k.x, correct? Because F=ma would apply to situation of uniform acceleration and acceleration isn't constant in this situation.

👎thumbup:?

 

[MedPR]

A 4- kg mass slides along a frictionless floor at 0.5 m/s until it collides with a spring. The mass comes to rest when the spring is compressed 5 cm. What is the force constant of the spring? Answer is 400 N/m.

Here's what I'm thinking.

To solve this problem, one ought NOT to use F=m.a=−k.x, correct? Because F=ma would apply to situation of uniform acceleration and acceleration isn't constant in this situation.

👎thumbup:?

How do you solve it?

 

[SaintJude]

Conservation of energy

Initially all energy is Kinetic energy. After mass comes to rest all the KE has been converted to potential elastic energy.

So KE=U

1/2mv^2 = 1/2 kx^2

 

[hopeful22213]

You do not need F=ma because this is an energy question. The block is sliding at a constant velocity so you worry about its kinetic energy. When it hits a spring and comes to a rest (for an instant, we assume) all kinetic energy has been converted to spring potential.

Kinetic energy is 1/2 mv^2
Spring potential is 1/2 kx^s where x is the distance it's compressed.

Setting them equal you have 1/2mv^2 = 1/2kx^2
(.5)(4)(.5)^2 = (.5)(k)(.05)^2
k=400

 

[hopeful22213]

Whoops. Hadn't refreshed the page before posting and missed your response St. Jude.

 

[MedPR]

Conservation of energy

Initially all energy is Kinetic energy. After mass comes to rest all the KE has been converted to potential elastic energy.

So KE=U

1/2mv^2 = 1/2 kx^2

Huh.

(.5)(4)(.25)(2)/(.0025)=k. Cool 🙂

Edit: Oops, missed x^2.

 

[SaintJude]

I'm asking if F=ma is VALID not if it's needed.

 

[scotchtapetest]

I'm asking if F=ma is VALID not if it's needed.

Are you asking if the most fundamental formula of physics is valid? 😕 I think you are asking if you could use that formula in this particular question....

You can't use F=ma, not because the acceleration (or deceleration in this case) is not constant but because you don't know what it is.... A non-constant acceleration can still be used in F=ma formula using integration...

 

[milski]

I'm asking if F=ma is VALID not if it's needed.

It's valid, but you need a formula for the force. The only way to get that is to put it in relation to the displacement of the object, which leads to a differential equation. The solution will lead you to the formulas for simple harmonic motion but approaching it as an energy problem is much more simple.

 

[MedPR]

I'm asking if F=ma is VALID not if it's needed.

Sure, if you know calculus 🙂

 

[SaintJude]

I don't think I originally expressed myself well. Please forgive me and I'll try again.

Say a student decided to use v^2=vo^2 + 2ax and then found that a.

And then plugged it into the F=-kx= ma. That would be incorrect. And now I seem to have answered my own question.

The reason this approach would be incorrect is not because the F=ma law, but b/c of previous use of one the "uniform acceleration" equations that might have been incorrectly used to find the a to plug into F=ma. Since acceleration isn't constant in this situation once the mass has hit the spring,using one of those uniform kinematic equations would be incorrect. (Yeah?)

 

[scotchtapetest]

I don't think I originally expressed myself well. Please forgive me and I'll try again.

Say a student decided to use v^2=vo^2 + 2ax and then found that a.

And then plugged it into the F=-kx= ma. That would be incorrect. And now I seem to have answered my own question.

The reason this approach would be incorrect is not because the F=ma law, but b/c of previous use of one the "uniform acceleration" equations that might have been incorrectly used to find the a to plug into F=ma. Since acceleration isn't constant in this situation once the mass has hit the spring,using one of those uniform kinematic equations would be incorrect. (Yeah?)

Did you read my post?

1) You can't find "a" that way, b/c "a" in this case would be a variable of time (i.e. as the spring is compressed more, the "a" would increase); The formula you mentioned for finding "a" only applies to constant acceleration/deceleration; "a" refers to deceleration in this case.

2) However, if you were able to find a(t) (or more correctly were given a(t) or came up with it based on experimental data) then you could use integration to solve for it using a(t) and F=ma.

 

[MedPR]

I don't think I originally expressed myself well. Please forgive me and I'll try again.

Say a student decided to use v^2=vo^2 + 2ax and then found that a.

And then plugged it into the F=-kx= ma. That would be incorrect. And now I seem to have answered my own question.

The reason this approach would be incorrect is not because the F=ma law, but b/c of previous use of one the "uniform acceleration" equations that might have been incorrectly used to find the a to plug into F=ma. Since acceleration isn't constant in this situation once the mass has hit the spring,using one of those uniform kinematic equations would be incorrect. (Yeah?)

As individuals that only know how to use F=ma for constant acceleration, you are correct.

For people who know how to integrate, you are incorrect.