“G’s” subjected to spring-mass system when dropped

[hootrpootr]

Hello,

I have an interesting spring energy problem that I’m trying to solve. I have a spring (known k) attached to a releasable hook at the top and a mass (known m) attaches to the bottom. This is a drop test, so when the releasable hook detaches, the spring and mass fall together a set distance (h). At that point, the spring engages and extends an unknown x.

I believe I know how to determine x using energy equations. Beginning energy is due to the height of the system (h+x) + the initial spring extension due to the mass hanging at rest. The end energy should be due to the spring energy of the system when it is fully extended.

Now finally, my question is: I can determine x, but how do I determine the “g’s” acting on the mass due to the spring? This is a requirement from the customer. Isn’t the g force just the acceleration/9.81m/s2?

Thanks! Hopefully this all makes sense.

 

[Tomfh]

G force = 9.8 is the gravitational force when the object is stationary, ie when you are sitting in a chair.

When the object is being pulled upwards by a spring this adds to the G force.

You need to work out the peak force the spring exerts on the object (spring constant multiplied by maximum extension) and convert that to acceleration. F=ma, etc

 

[r13]

I think so, because you convert potential energy into kinetic energy with constant "g", but varying "a", IMO.

 

[3DDave]

The maximum acceleration will be at the maximum spring extension where the spring will have stored the potential energy of the fall into the potential energy of the spring extension.

 

[IRstuff]

F = -k * x

U = 1/2 * k * x^2

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[hootrpootr]

I think I agree with all the comments so far. Thanks for the responses!

Just to clarify, once I’ve calculated the extension of the spring and then write F=ma, it’s really just a simple FBD- so: Fspring - mg = ma. And then I’d solve for “a”?

 

[r13]

The linked page can help. Pay attention to "acceleration" and "free fall". Link

 

[Tomfh]

Hoot, I think you’re right. At peak extension the object is stationary. There is the load from the spring pulling the object up, and its own weight pulling it down. That leaves a net force, which when divided by the objects mass gives you the peak upwards acceleration, which you divide by 9.8 to get G force.

 

[IRstuff]

Technically, dividing 9.80665 m/s^2 gives you g's of acceleration. Note that "G" is the symbol for the universal gravitational constant ala Newton: 6.67 x 10-11 m^3 kg^-1 s^-2

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[r13]

I think below is what the OP is looking for.

Let the free fall distance just before engage the spring = h, and the length of the stretch = s

- Find velocity at the end of the free fall (spring is yet to be engaged), from conservation of energy:
m*g*h = m*V²/2 -----> V = √2*g*h
- Find acceleration during the stretch of the spring, Vi = V, Vf =0
a = (Vi² - Vf²)/(2*s) -----> a = g*(h/s)

∝ = a/g = h/s

 

[rb1957]

I guess I'd approach the problem differently.

at the bottom of the mass's travel, with the spring extended "s" the acceleration of the body is k*s/m-g, or "g"s is k*s/(m*g)-1.

another day in paradise, or is paradise one day closer ?

 

[HTURKAK]

..........Now finally, my question is: I can determine x, but how do I determine the “g’s” acting on the mass due to the spring? This is a requirement from the customer. Isn’t the g force just the acceleration/9.81m/s2?....

Correct me if i understand the problem wrong= a spring with ( k ) is attached to releasable hook at level of mass is H , and there is a mass m attached to the bottom of spring with k attached to a releasable hook at the top and a mass (known m) attaches to the bottom. The releasable hook detaches, the spring and mass fall together a set distance (h) and the spring engages and extends an unknown A. and the question is, What is the acceleration of mass at any position?

Assuming the mass of spring is negligible and undamped Free Vibration, the motion will be Simple harmonic motion.
With energy approach, knowing the total energy is conctant, and X is the distance of mass at any time after the spring engaged,
Velocity of mass when the spring engaged Vo=sqrt (2gh)
KE of mass whenthe spring engaged KE =(1/2 )M*Vo**2

And remaining of calculation i worked on notebook and picture is below: