Finding acceleration of two masses connected by spring horizontally

  • Thread starter Revivalist
  • Start date
  • #1

Homework Statement


Two masses sit on a horizontal surface and are connected by spring with known spring constant k. There is a known coefficient of kinetic friction µ_k for both blocks. The masses are brought together so that the spring is compressed by an amount Δx and then released. Find the acceleration of each block when they start moving.


Homework Equations


Hook's Law: F=-kΔx
Newton's Second Law: F=ma
Friction force: F=µ_k*N


The Attempt at a Solution


I originally tried to solve the problem by just doing a summation of forces on each block (m1 and m2) using Hook's Law.

ΣF=kΔx - µ_k*m1*g = m1*a1

ΣF=kΔx - µ_k*m2*g = m2*a2

But this simply gives a1 = (kΔx - µ_k*m1*g)/m1 and a similar equation for a2.

However, I realized that the problem with this is that the equation for a1 is treating m2 as a fixed body. But since m2 is going to accelerate in the other direction, this should have an effect on the acceleration of m1 by decreasing the force it experiences from the spring. In fact, I was thinking that the equation should reduce to what I have above only when m2 is vastly greater than m1 because that means m2 would be like a rigid wall compared to m1. Then the equation above should properly describe what is happening to m1.

Likewise, if m1 is vastly greater than m2, I should find that m1 does not accelerate at all and m2 is the only that accelerates. So I'm obviously neglecting the fact that because both masses are accelerating, this needs to be incorporated into the accerlation equations somehow.

I also became confused when I realized that if the blocks are moving then that means the acceleration will actually be changing sinusoidally. However, I am going to assume that the question is asking about the acceleration right at the moment that the blocks start to move. Otherwise, we would have to use a sinsoidal expression for the acceleration and that would be dependant on time.

Any help would be greatly appreciatd. Thanks!
 

Answers and Replies

  • #2
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,133
160
But this simply gives a1 = (kΔx - µ_k*m1*g)/m1 and a similar equation for a2.

I think that's right.

The fact that the other mass also moves will affect the value of Δx at later times. However, they just want the initial acceleration, and Δx is a given, so what you wrote should be sufficient.

p.s.
Welcome to PF :smile:
 
  • #3
Doc Al
Mentor
45,140
1,442

The Attempt at a Solution


I originally tried to solve the problem by just doing a summation of forces on each block (m1 and m2) using Hook's Law.

ΣF=kΔx - µ_k*m1*g = m1*a1

ΣF=kΔx - µ_k*m2*g = m2*a2

But this simply gives a1 = (kΔx - µ_k*m1*g)/m1 and a similar equation for a2.
Looks good to me. (Assuming, of course, that the spring force is greater than the friction.)

However, I realized that the problem with this is that the equation for a1 is treating m2 as a fixed body. But since m2 is going to accelerate in the other direction, this should have an effect on the acceleration of m1 by decreasing the force it experiences from the spring.
Not really. At any given instant, the force exerted by the spring on each mass depends only on the stretch of the spring.
In fact, I was thinking that the equation should reduce to what I have above only when m2 is vastly greater than m1 because that means m2 would be like a rigid wall compared to m1. Then the equation above should properly describe what is happening to m1.
All you are asked to find is the acceleration at one moment, which you've done. But you are correct that the ultimate motion of each mass depends on the relative size of their masses.
 

Related Threads on Finding acceleration of two masses connected by spring horizontally

  • Last Post
Replies
2
Views
3K
Replies
11
Views
21K
Replies
18
Views
270
Replies
4
Views
923
Replies
9
Views
6K
  • Last Post
Replies
6
Views
20K
  • Last Post
Replies
4
Views
5K
Replies
2
Views
2K
Replies
2
Views
10K
  • Last Post
Replies
1
Views
10K
Top