# Block dropped on a spring

1. Mar 1, 2006

### scavok

A 2.8 kg block is dropped from a height of 5.8 m (above the top of the spring) onto a spring of spring constant 3955 N/m. Find the speed of the block when the compression of the spring is 15.0 cm.

x=0.15m
k=3955N/m
m=2.8kg

Einitial mech=Efinal mech

Usi+Ugi+Ki=Usf+Ugf+Kf

Usi=0 because the spring is uncompressed
Ugi=???
Ki=.5mvi2 where $$V_i =sqrt(2g*5.8m)=10.868$$

Usf=.5kx2
Ugf=0 because I can set the height at that point to be 0
Kf=.5mvf2 where we're solving for v

How do I calculate the potential graviational energy? I'm guessing I need to find out how far the block would go if it were allowed to come to a rest, but I get an impossible to solve equation, or maybe I just don't remember what to do.

Block at rest:
Usi+Ugi+Ki=Usf+Ugf+Kf

0+mgx + .5mv2i=.5kx2
x(mg-.5kx)=-.5mvi2
x(27.468-1977.5x)=-165.359
x=???

Last edited: Mar 1, 2006
2. Mar 1, 2006

### DaMastaofFisix

Hey dude, no worries. Though it might seem tacky with the math, you might wanna set you gravitational potential energy to be zero at the equilibrium point of the block-spring system. It's a bit of extra math, but it takes all of two second to calulate. If not, just set the zero point at the point of contact with the springand ignore changes in potential energy (though I could be worng in that assumption). You might wanna rethink this a little bit, but you're on the right track. Just realize tha the mgh at the beginning is all converted when the spring is compressed at .15 meters. In other words, mgh=.5kx^2+.5mv^2, where h is the height above the spring, and x is the compressed length of the spring. Good luck with it dude!

3. Mar 1, 2006

### scavok

Sorry, I still don't have a clue.

4. Mar 2, 2006

### scavok

Just a bump. I finished all my other problems, but I still can't picture what I need to do for this.

5. Mar 2, 2006

### vaishakh

The body is using the stored graavitational potential energy to do the compression on the string. When the body falls, it loses some amount of potential enegy which gets converted to Kinetic energy, i.e. 1/2mv^2 = mgh.
Then this Kinetic enrgy has the impact on spring. Find the work done by the box on the spring when it does the given work on the spring. It loses that much Kietic energy. But gains Kinetic energy by the conversion of potential energy from the height reference level of natural state of spring.

IF YOU DON'T KNOW THE POTENTIAL ENERGY, ONLY CALCULATE THE CHANGE IN THEM.