How can I calculate the speed of a block dropped on a spring?

  • Thread starter Thread starter scavok
  • Start date Start date
  • Tags Tags
    Block Spring
AI Thread Summary
To calculate the speed of a block dropped onto a spring, the conservation of mechanical energy principle is applied. The initial gravitational potential energy of the block is converted into kinetic energy and the potential energy stored in the spring when compressed. The equations used include the gravitational potential energy (mgh), kinetic energy (0.5mv^2), and spring potential energy (0.5kx^2). Setting the gravitational potential energy to zero at the equilibrium point simplifies calculations. Ultimately, the relationship between these energies allows for the determination of the block's speed at the point of maximum spring compression.
scavok
Messages
26
Reaction score
0
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:
Physics news on Phys.org
Hey dude, no worries. Though it might seem tacky with the math, you might want to 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 want to 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!
 
Sorry, I still don't have a clue.
 
Just a bump. I finished all my other problems, but I still can't picture what I need to do for this.
 
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 energy 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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top