# Mass on a spring inside a box falling under gravity

1. Mar 1, 2009

### naggy

1. The problem statement, all variables and given/known data

A mass m is attached to a spring(massless) that is located inside a massless box. The box is falling under gravity. When the box starts to fall the spring is in it's equilibrium position and the box sticks to the ground when it hits it.

-The box is a distance H from the ground
-Spring has spring constant k
-The mass on the spring is m

Find the equation of motion (and initial conditions) when
a)the box is falling and
b)when the box has landed.

Variables
x is movement from equilibrium position of spring
y is distance from ground to mass

2. Relevant equations

$$L=KE - PE$$
or
$$F=m\ddot{x}$$

3. The attempt at a solution
I prefer using Lagrangian equations. When the box is falling:
$$KE= \frac{1}{2}m\dot{x^2}$$
$$PE= mgy +\frac{1}{2}kx^2$$

Now can I connect y(distance from the ground to m) and x(movement from equilibrium position of mass) with y=constant + x and use the Euler lagrange equations?

I'm also not sure on intial conditions, it would be x(0)=0 and x'(0)=0 for the first eq. of motion

when the box lands, maybe x(tH)=H and x'(tH)=sqrt(2gH) ??

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Last edited: Mar 1, 2009
2. Mar 2, 2009

### lanedance

i think you're on the right track.. i would spilt it up to look at before & after impact separately

so define
x - spring extension
y - distance base of spring zero to ground
then z = x + y is distance from ground to mass

so say time of impact is t0

before landing everthing is accelerating downward at g
so
x(t<t0)=0
x'(t<t0)=0
and
PE = my'^{2}/2
KE = mgy

directly before impact find
v0 = z'(t0) = y'(t0)

then inatantly after impact the box no longer moves
y(t>t0) = 0
y'(t>t0) = 0

and the intial conditions of the mass are
x(t0) = 0
x'(t0) = v0

3. Mar 3, 2009

### naggy

what about the Potential energy of the spring?

And what is base of spring? Y is distance base to spring? Do you mean the springs equilibrium position?

4. Mar 3, 2009

### lanedance

hey naggy

the way i set it up y was the springs equilibirum position, howvere you can set it up howvver you like. i didn't write down the final energy equations, so you need to include the PE of the spring & gravitational PE in the final section

the main thing to realise is beofre landing everything will have constant downward acceleration of g.

when the box strikes it will have intially total energy in the form of KE of the spring. this will then oscillate between spring PE & mass KE, with some gravitational PE.

you can make life easier for yourself by choosing a spring equilibrium position that essentially removes the complication of gravity PE (where does the spring sit under gravitational force of math)

5. Mar 4, 2009

### naggy

something that is bothering me

When the the box is falling:

$$\ddot{z}=-g$$ but I don´t know the initial conditions here? z=x+y but I don´t know the height up to the mass I only know the height to the bottom of the box?

6. Mar 4, 2009

### naggy

But does the spring oscillate at all during the flight down?

7. Mar 4, 2009

### lanedance

I would say no, in an idealised free fall acceleration of the box spring and mass will all be identical under gravity, so there will be nothing to cause relative motion, that is until the box lands & sticks

unless you are given something different in the initial conditions, this is how I would approach the problem