Compression of a spring by an object in free fall

[hawkeye1029]

Homework Statement

A 2.0 kg object starting at rest falls 3.0 m onto a 4.0 N/m spring. How far will the spring be compressed?

m = 2.0 kg
v_o = 0 m/s²
y = 3 m
k = 4 N/m
g = -9/8 m/s²

Homework Equations

eq1: Fg = mg
eq2: F_spring = -kx

The Attempt at a Solution

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eq1: F_g = mg
F_g = (2.0kg)(9.8m/s²)
F_g = 19.6N

F_spring = -kx
19.6N = (-4N/m)x
x = -4.9m -> compressed 4.9m

Since they gave y = 3m, I feel like I need to use it, but I'm not sure where. The answer key says the correct answer is 5.42m.

Thank you for any help.

 

[BvU]

I feel like I need to use it, but I'm not sure where

Well, you can kind of imagine the amount of compression depends on the height from which the object is dropped ...

What you describe is placing the object on the spring and gently lower it until equilibrium is reached.

 

[gneill]

The answer key says the correct answer is 5.42m

Looks to me like the answer key didn't account for the additional GPE converted as the spring compresses.

 

[hawkeye1029]

Well, you can kind of imagine the amount of compression depends on the height from which the object is dropped ...

What you describe is placing the object on the spring and gently lower it until equilibrium is reached.

OK, so then I would need to use Conservation of Energy? Potential energy of block -> kinetic energy -> potential energy in spring?

mgh = 1/2kx²
x² = 2mgh/k
x = 5.42m

Looks to me like the answer key didn't account for the additional GPE converted as the spring compresses.

Do you mean the GPE of the block when it's resting on the spring?

 

[RPinPA]

OK, so then I would need to use Conservation of Energy? Potential energy of block -> kinetic energy -> potential energy in spring?

mgh = 1/2kx²
x² = 2mgh/k
x = 5.42m
Do you mean the GPE of the block when it's resting on the spring?

Yes, energy is the easiest way to solve a problem like this. It gains kinetic energy from the GPE of falling 3.0 meters plus the additional distance $x$ as it compresses the spring. All of that energy is converted into spring compression energy.

As @gneill says above, it looks like the answer key didn't account for the additional energy $mgx$ from falling that extra distance $x$.

 

[gneill]

Do you mean the GPE of the block when it's resting on the spring?

No, I mean the GPE that is converted to KE (and hence to spring PE) while the block is still dropping and compressing the spring. The spring doesn't just fall through 3 m. That 3 m fall just brings it into first contact with the spring. As the spring compresses, the block continues to drop along with the top of the spring.

 

[hawkeye1029]

No, I mean the GPE that is converted to KE (and hence to spring PE) while the block is still dropping and compressing the spring. The spring doesn't just fall through 3 m. That 3 m fall just brings it into first contact with the spring. As the spring compresses, the block continues to drop along with the top of the spring.

Ohh I see. As I'm in beginning physics and this is a problem at the start of a new unit, I think the teacher didn't want to confuse us.
If I did want to take that into account, how would I go about doing that?

 

[gneill]

Ohh I see. As I'm in beginning physics and this is a problem at the start of a new unit, I think the teacher didn't want to confuse us.
If I did want to take that into account, how would I go about doing that?

It would be a lot more confusing if you were starting out with incorrect information and having to un-learn it later

The ball falls an extra distance equal to the compression distance of the spring. It's the same x that you're looking for.

 

[RPinPA]

Ohh I see. As I'm in beginning physics and this is a problem at the start of a new unit, I think the teacher didn't want to confuse us.
If I did want to take that into account, how would I go about doing that?

See my answer. Add a term on the left to account for the object falling an extra distance $x$.