Maximum Deformation of Springs in a Collision: Deriving the Equations

[sumeer@dinsum.c]

Homework Statement

A block of mass 20kg with a spring with a stiffness of 500N/m attached to the bottom is dropped .5m from rest onto another spring of stiffness 800N/m. The weight of the springs can be neglected. What is the maximum deformation of each spring due to the collision.

Homework Equations

work = change in kinetic energy + change in potential energy + change in elastic energy

The Attempt at a Solution

I have assumed that I can add the different spring coefficients, giving me an energy equation of:

work=0= -.5mv^2 + -mg(.5+x) + .5k(x^2)

here k is the sum of the spring coefficents and x is the total deformation of the spring.
I solved the equation, got two answers for x one positive and one negative.

Can I use the ratio of the spring coefficients to work out the individual reformation, and also do i use the positive value calculated for x?

Sorry for the long question, your help is appreciated.

 

[BerryBoy]

Ermm, if the springs are inline with each other (on top of each other) you add the reciprocals of the constants, so that:

{k_T}^{-1} = {k_1}^{-1} + {k_2}^{-1}

The system adds (becomes stiffer) when the springs are used adjacent to each other.

So I think, from the wording of the question, you need to use this equation maybe:

k_T = \frac{k_1 k_2}{k_1 + k_2} - Your thoughts

But yeah - the kinetic energy term would become mg\Delta h in your equation (because the kinetic energy came about from falling the distance \Delta h, cause when it hits the spring, your mgh = 0. Apart from that, logic seems fine...

Let me know how you get on :P
Sam

 

[saket]

I do agree with Sam.. Keq = k.k'/(k + k') , is the one to be used.
Since, the two springs are adjacent, force in them must be equal => k.x = k'.x'
Equivalent spring, Keq.X = k.x = k'.x', where X = x + x'.

@ sumeer@dinsum.c
I posted this, just to urge one thing: please REASON OUT your assumptions! Do not make blind assumptions. The assumptions you made (that Keq = k+ k') would have been true, if elongations in both the springs would have been same! x = x', X = x = x', and Ftot = Keq.X = k.x + k'.x'
I hope you get the idea.

 

[sumeer@dinsum.c]

Hi Sam,

Thanks for your information, I've managed to work through the problem.
Sumeer

 

[rl.bhat]

Hi Sameer, I don't how you solved the problem. But see whether it matches with this. Loss of PE of block = mg(0.5 +X). This energy is shared by the two springs and it is equal to 1/2*kx^2 + 1/2*k'x'^2. During the compression force in each spring is kx and k'x'. Hence down ward force =mg + kx and upward force = k'x'. In equilibrium position mg +kx = k'x' and X = x+x', write x' and X in terms of x and substitute it in the above energy equation. Then solve for x, and x'