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

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AI Thread Summary
To determine the applied force on the crates connected by a spring, the spring force formula Fspring = -k*x is used, where k is the spring constant and x is the compression distance. Given k = 8.1 kN/m and x = 0.051 m, the spring force calculates to 0.4131 kN. The differing masses of the crates do not affect the calculation of the spring force directly; however, they may be relevant for understanding the overall system dynamics. The applied force must equal the spring force to maintain equilibrium, resulting in an applied force of approximately 0.4131 kN. Thus, the solution confirms that the spring's compression directly provides the necessary force without needing to factor in the masses.
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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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