physics problem with springs

[vsl5]

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Two masses (650kg and 490kg) are attached by a massless spring (k=8000N/m) and lie horizontally. A constant force F is applied to mass 1 (650kg mass) which ultimately gives both masses and the spring a common acceleration on a frictionless surface. The spring is compressed 5cm from its initial unstretched length during the motion. What is force F?

A: 931N

Why is not simply F=k delta x since there is no friction and gravity is cancelled out by normal force? Thanks in advance.

 

[GRod18]

I'll take a crack at it, I don't know if you have solved this problem, but to me its a pretty complicated hooke's law problem, since Rabolisk and Fizzgig have not taken a crack at I will go ahead.

So since your mass is accelerating, you know that there is a net force. The force on your smaller mass 490 Kg would be equal and opposite the spring force which you find via hooke's law equation to be 400 N. So we have 400N acting on 490 Kg. Because we have a constant force being applied, same amount of force should spread evenly across all mass and spring. So you can find the acceleration to be 400N/490Kg = .82 m/s^2 on the smaller mass, this same acceleration would act on the bigger mass, giving the net force on the bigger mass to be F= ma = 530N. you then add 530N+400N to get your final value of 930N. This is the total force applied.

Maybe Rabolisk and Fizzgig as well as others might share some insight on this problem