Okay, so this is probably very easy, just checking the solution

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SUMMARY

The problem involves two crates with masses of 640 kg and 490 kg connected by a massless spring with a spring constant of 8.1 kN/m. When the spring compresses by 5.1 cm, the applied force can be calculated using Hooke's Law, Fspring = -k*x. The correct calculation is F = 8.1 kN/m * 0.051 m, resulting in an applied force of 0.4131 kN. The masses of the crates do not affect the calculation of the spring force in this scenario.

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Homework Statement


Okay guys, I'm new here, and this problem just annoyed me because I believe it's very simple, but it looks like it could be more complicated.

Two large crates, with masses 640 kg and 490 kg, are connected by a stiff, massless spring (k=8.1 kN/m) and propelled along an essentially frictionless, level factory floor by a force applied horizontally to the more massive crate. If the spring compresses 5.1 cm from its equilibrium length, what is the applied force?


Homework Equations



I think it's just this one: Fspring = -k*x


The Attempt at a Solution



Okay, the only thing that makes it sound at all complicated is that the crates are different masses, and that they talk about mass at all. But I want to say it's just 8.1*(.051) to get kN. Is that it? Or should I take into account the masses of the boxes. In which case I'm not quite sure where to go.
 
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