Two masses attached by a spring

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Homework Help Overview

The problem involves two masses connected by a spring, with a force applied to one of the masses. The context is to determine the acceleration of the system, which includes understanding the forces acting on each mass and the role of the spring.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to analyze the forces acting on both masses and equate their accelerations, while also questioning the correctness of their approach compared to a classmate's method.
  • Some participants question the application of the force and its impact on the two masses, particularly regarding the nature of the forces acting on the 2 kg mass.

Discussion Status

The discussion is exploring the forces acting on the two masses, with some guidance provided regarding the nature of contact forces versus action-at-a-distance forces. Participants are clarifying their understanding of the net forces involved.

Contextual Notes

There is a focus on the distinction between forces acting on the two masses, particularly the applied force and the spring force, as well as the implications of these forces on the net force calculations for each mass.

fuddyduddy
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This problem was on my test today and I am not sure if my solution is correct.

Homework Statement


A box of mass 2 kg and another of mass 8 kg are attached by a spring with a spring constant of 80 N/m. A 4.0 N force is applied to the 8 kg box. Both of the boxes move with a constant acceleration on a horizontal frictionless surface. Find the acceleration.


Homework Equations


Fs = -kx
Fnet = ma

The Attempt at a Solution


Here is where I'm not sure if I am right or if my classmate is right. He did this:

total mass of system = 10 kg, total applied force = 4.0 N
4 = (10 kg)(a)
a = 0.4

What I did was find the Fnet of the 8 kg box and the 2 kg box respectively and set the accelerations as equal. I think this is where I kind of went off the rails - I pictured it as the spring pulling in on both of the boxes, meaning the Fnet of the 8 kg box is 4 - Fs and the Fnet for the 2 kg is 4 + Fs.

Box A (2 kg)
Fnet = ma
a = Fnet/m
a = (4 + Fs)/2

Box B (8 kg)
a = Fnet/m
a = (4 - Fs)/8

Then set them equal: (4 + Fs)/2 = (4 - Fs)/8

Cross multiply to get: 32 + 8Fs = 8 - 2Fs
10Fs = -24
Fs = -2.4 N

When I plug in these numbers into the free body diagrams it all seems to work out, but that might be circular logic - I get an acceleration of 0.8 m/s^2

Please help! :(
 
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Hello fuddyduddy. Welcome to PF!

fuddyduddy said:
Box A (2 kg)
Fnet = ma
a = Fnet/m
a = (4 + Fs)/2

Does the 4 N force act on the 2 kg mass or does it act only on the 8 kg mass?
 
It acts on the 8 kg mass, the picture was as such:

[2 kg]--(spring)--[8 kg]--> 4.0 N

And thank you for the welcome! :)
 
OK. When thinking about the forces acting on on the 2 kg mass, the force of gravity is an "action-at-a-distance" force (from the earth). All other forces on the 2 kg mass are "contact" forces coming from other objects in contact with the 2 kg mass.

So, if someone pulls on the 8 kg mass to create the 4.0 N force on the 8 kg mass, that pulling force is a contact force on the 8 kg mass, not a force on the 2 kg mass.

So, what should you write for the net horizontal force on the 2 kg mass?
 
Okay, I see now! So the net force acting on the 2 kg block is simply the spring force directed to the right. Thank you!
 
Yes. Good.
 

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