Two masses attached by a spring

  • Thread starter Thread starter fuddyduddy
  • Start date Start date
  • Tags Tags
    Spring Two masses
AI Thread Summary
The problem involves two masses, 2 kg and 8 kg, connected by a spring with a spring constant of 80 N/m, and a 4.0 N force applied to the 8 kg mass. A classmate calculated the acceleration as 0.4 m/s² by using the total mass and applied force, while the original poster attempted a more complex approach involving net forces on both masses. The correct interpretation is that the 4 N force acts only on the 8 kg mass, and the net force on the 2 kg mass is solely due to the spring force. This clarification leads to the conclusion that the net force on the 2 kg mass is determined by the spring, not the applied force. Understanding the forces acting on each mass is crucial for solving the problem accurately.
fuddyduddy
Messages
13
Reaction score
0
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! :(
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top