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/sci/ - Science & Math


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2286974 No.2286974 [Reply] [Original]

Hey /sci/

If a spring with a spring constant of 10 N/m is compressed by a 3 kg block of whatever 12 meters, what's the speed at the end of the 12 meters? It doesn't apply 120 N the full 12 meters, does it?

>> No.2286982

bump

>> No.2286998

>metric system
fuck off

>> No.2286999

Use conservation of energy.

>> No.2287009

>>2286974
use Kinetic energy = potential energy

>> No.2287034

Pro tip: You don't post a question on an international board using specific units. You use "spring constant k, mass m, distance s".

We cannot answer the question. There's probably a maximum velocity that the spring can extend at, such that if the block reaches this velocity, no further acceleration occurs. Other than that, it appears you are working under the assumption that springs follow Hooke's law, in which case you can compute the force by applying that exact law.

>> No.2287038
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2287038

>>2286998

>> No.2287055

Itt, fags who compute their problems using numbers instead of letters.

>> No.2287056

>>2286999
agreed
Write the expression of the total energy of the system (potential energy and kinetic energy) for the initial state and the final state. Those two expressions should be equal if there is no loss of energy (i.e. no friction etc.).
You should then be able to isolate the speed from those two expressions.

>> No.2287065
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2287065

Elastic potential energy = 1/2*K*X^2

PE = KE

KE= 1/2*M*V^2

Solve for V

>>2287034
>>2286998
>>2287055

Pic related

>> No.2287074

You've probably met the harmonic oscillator, so you can easily solve this problem by applying that solution:
<div class="math">m \ddot{x} = -kx,</div>
with your starting conditions, the solution to this is
<div class="math">x=A \sin (\omega t),</div>
where A is the starting amplitude, <span class="math">\omega = \sqrt{k/m}[/spoiler]. Then
<div class="math">\dot{x}=\omega A \sin(\omega t),</div>
and the block obviously let's go when the velocity reaches it's peak, in this case <span class="math">\dot{x} = \omega A[/spoiler].

Or you could use energy conservation
<div class="math">E=1/2 m \omega ^2 A^2 = 1/2 m \dot{x} ^2,</div>
which gives the same result

>> No.2287085

>>2287074

You forgot you're cool face

Op's question is a simple physics 101 question. Schools don't even teach harmonics in 101 level classes

>> No.2287087

>>2287074
hurr

>> No.2287097

>>2287085
They should.

>> No.2287107

>>2287097

Most students can barely comprehend conservation of energy problems like this one.

You wold be amazed at how few people can do the "dreaded" incline plane with friction problem.

The only requirement for physics at most universities now is algebra 1

>> No.2287116

>>2287085
weird I did learn about the harmonic oscillator the first year...
europefag here

>> No.2287120

>>2287107
Then what's the point of a question such as this, if not to demonstrate that the two solutions are equivalent?

>> No.2287127

>>2287120

What I wrote here >>2287065 is about all the explanation they will give in state schools for how to solve this problem.

"All you need to know is PE=KE" essentially sums in up.

This wasn't just my school either, the people who I have tutored from other schools have this same problem

>> No.2287154

>>2287127
Wow, that's retarded... Reminds me of this teacher I once met who asked me "Can you explain to me just why work equals force times distance? It's so intuitive to me, but I have a hard time explaining people why it's like that."

mhmm.