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Acceleration by a spring...


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#1 Externet

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Posted 5 January 2017 - 05:25 AM

|\/\/\/\/\/\/\/\/\/\/\|========================    --->

 

 

Took 356 Newtons to compress a spring 0.5 metre.  A 1 Kg rod contacting its end gets pushed when the spring is released.

Is that enough data to calculate the acceleration, or :

 

Took 356 Newtons to compress a spring.  A 1 Kg rod contacting its end gets pushed during 0.1 second when the spring is released to full extension.

Is that enough data to calculate the acceleration ?

 

What formula should be applied ? -Ignoring friction, drag-

Is the calculated acceleration result the one at the end of spring extension; at position 0.5m away and 0.1 second from firing ?

 

( The initial force applied to the rod is 356 Newtons and linearly? decreases to zero in 0.5m of travel that takes 0.1 second. )

 

Nope, not homework :-)

 

 

 

 

 

 

 


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#2 J.C.MacSwell

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Posted 5 January 2017 - 07:53 AM

You have enough if you include enough simplifying assumptions, Ideal spring, massless, force is proportional to displacement.

 

The acceleration can be calculated at any point after release and will be proportional to the force and thus displacement where a=F/m. 

 

You can use calculus to do the same wrt time.


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#3 Sriman Dutta

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Posted 5 January 2017 - 11:48 AM

I think.....

a=F/m=kx/m


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#4 Externet

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Posted 5 January 2017 - 03:41 PM

 

Please check my rationale...

 

Initial velocity Vi is 0, spring compressed.
Impulse force F applied is 356 Newtons at start of travel.
Time the force is applied is 0.1 seconds during spring expansion = Δt
Final velocity Vf at release is unknown = Vf = Vfinal - Vinitial = Δv

Acceleration is a = Δv/Δt = F/m
Δv = Δt F /m = 0.1s x 356N / 1Kg = 35.6 m/s <---- take-off velocity.

a = 35.6m/s / 0.1s = 356m/s^2

The kinetic energy Ek =  mv^2 / 2
1 x 35.6 x 35.6 /2 = 633.68 Joules.

 

But, do I instead need to integrate the force applied during 0.1 second while linearly decreases to zero ?


Edited by Externet, 5 January 2017 - 03:52 PM.

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#5 Janus

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Posted 5 January 2017 - 04:55 PM

The acceleration will not be constant as the spring depresses. It will decrease from 356 m/s2 to 0 m/s2 over the 1/2 meter.  So I assume you are looking for the net acceleration.  In which case you have to integrate the force over the 1/2 meter to get the energy difference between start and finish, and then use this to determine the final speed of the rod.


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#6 Externet

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Posted 5 January 2017 - 08:56 PM

Thanks, Janus.

Then, there is where am stuck.  To express the event into an integral equation.

 

Potential energy of the compressed spring Ep = 1/2 x F x d = 356N x 0.5m / 2  = 89 J.

Which with 'no losses', transfers to the rod as kinetic energy Ek:

 

Ep = Ek

 

Ek = 89 J = 1/2 x m x v^2

 

v^2 = 89 J x 2 / 1Kg

 

v = 13.34 m/s

 

Got it or goofed ?


Edited by Externet, 5 January 2017 - 10:30 PM.

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#7 Bender

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Posted 6 January 2017 - 05:35 PM

Looks correct. Much easier that way than doing the integration, although the result will be the same.


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#8 Externet

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Posted 8 January 2017 - 11:49 PM

I thank you Bender very much for replying,


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#9 zztop

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Posted 9 January 2017 - 12:00 AM

 


 

But, do I instead need to integrate the force applied during 0.1 second while linearly decreases to zero ?

 

You needed to form the differential equation:

 

m \frac{d^2x}{dt^2}=kx

 

Do you know how to set the initial conditions?

Do you know how to solve differential equations?


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#10 Externet

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Posted 9 January 2017 - 12:14 AM

Do you know how to set the initial conditions?

 No, and thank you.

 

Do you know how to solve differential equations?

I did, 44 years ago.

 


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#11 zztop

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Posted 9 January 2017 - 12:20 AM

 No, and thank you.

 

I did, 44 years ago.

 

The solution is x(t)=A e^{-t \sqrt{a}}+B e^{t \sqrt{a}}

 

where a=\frac{k}{m}

 

You can get A and B from setting the initial conditions. You have them in the problem.


Edited by zztop, 9 January 2017 - 12:21 AM.

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