Energy and Springs

[healthyeater]

A spring is compressed vertically 10 cm, and a 5 kg block is placed on top of it. The spring has a spring constant of 50 N/m. If the system is released from rest, what is the maximum height achieved by the block?


What I thought to do was to apply the conservation of energy and set initial conditions equal to final.

1/2kx^2=mg(h+x) and then plug in the values and solve.. I did not get the correct answer. Where did I go wrong?

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[image187]

"(h-x)" ? not sure but maybe? if the object gets released from the spring 10 cm off the ground, when you solve for h, what you're getting is the final height not account for the height it started at, so you'd have to end up adding this to the other side of the equation. am I right? or was my mcat way too long ago

 

[Captain Sisko]

A spring is compressed vertically 10 cm, and a 5 kg block is placed on top of it. The spring has a spring constant of 50 N/m. If the system is released from rest, what is the maximum height achieved by the block?


What I thought to do was to apply the conservation of energy and set initial conditions equal to final.

1/2kx^2=mg(h+x) and then plug in the values and solve.. I did not get the correct answer. Where did I go wrong?

Check and make sure your spring is stiff enough to actually lift the block. F =kx where x is. 1 so spring force is 5 n. Block weighs 50 n so it should go down, I think another 90 cm before it reaches equilibrium?

 

[LoLCareerGoals]

Check and make sure your spring is stiff enough to actually lift the block. F =kx where x is. 1 so spring force is 5 n. Block weighs 50 n so it should go down, I think another 90 cm before it reaches equilibrium?

Exactly. I was like....eh negative 90cm???

 

[Balantidiumcoli]

I think the other people got it all ready.

Force of the spring: kx = -.5 N = F1
Weight of mass: Mg = 50 N = F2
This means that the spring should be compressed further.

Equilibrium -> F1+F2=0 -> F1 = -50 N
Set -50 N = kx -> x=-1.0 m. Since the spring all was all ready compressed 10 cm that means it should compress by another 90cm.

 

[Captain Sisko]

It could settle in some sort of harmonic equilibrium, don't forget. When it settles, it has some velocity so it'll jiggle.

 

[Balantidiumcoli]

It could settle in some sort of harmonic equilibrium, don't forget. When it settles, it has some velocity so it'll jiggle.

Hmm... You're right. Since the spring was all ready compressed, it would allow the mass to lower past the equilibrium point.

(50)1 - (50).9 = 5 N of excess force past equilibrium -> -55N = kx -> x=-1.1m at the lowest point -> PE =.5kx^2=30.25 J. Set 30.25=mgh -> h= 0.6 m, h-.1m =0.5m past equilibrium or 40 cm lower than the starting point.

That right?

- assumes g=10m/s^2 -

 

[nb07013]

set PE of spring equal to PE of Object (The energy goes PE of spring--> KE of object --> PE of object at max height). Therefore, 1/2kx^2=mgh solve for h. h=0.005 m (which is the correct answer in the book). You do not need to minus the compression of the spring because the spring is already compressed when the object is placed!