Finding acceleration of two masses connected by spring horizontally

In summary: You'll probably learn more about this in your studies of dynamics.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.No, you've got it right. In fact, if m1 is much greater than m2, we know that the acceleration of m1 will be zero, because the spring will be pulling with a force that is much greater than the friction force. If m2 is much greater than m1, then the situation is just the opposite. But for
  • #1
Revivalist
1
0

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!
 
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  • #2
Revivalist said:
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
Revivalist said:

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.
 

1. What is the equation for finding acceleration in this scenario?

The equation for finding acceleration in this scenario is a = (F - kx)/m, where a is the acceleration, F is the applied force, k is the spring constant, x is the displacement of the spring, and m is the mass of the object.

2. How do you calculate the spring constant?

The spring constant can be calculated by dividing the applied force by the displacement of the spring. This can be written as k = F/x.

3. Can you explain the relationship between mass and acceleration in this scenario?

The relationship between mass and acceleration in this scenario is inverse. This means that as the mass increases, the acceleration decreases. This is because a heavier object requires more force to accelerate at the same rate as a lighter object.

4. How does the initial displacement of the spring affect the acceleration?

The initial displacement of the spring does not have a significant effect on the acceleration. This is because the initial displacement is usually very small compared to the displacement caused by the applied force. However, it can affect the amplitude of the resulting oscillations.

5. Are there any assumptions made when calculating acceleration in this scenario?

Yes, there are a few assumptions made when calculating acceleration in this scenario. Firstly, it assumes that the spring is ideal and obeys Hooke's law. It also assumes that there is no friction or air resistance acting on the masses. Additionally, it assumes that the masses are connected by a rigid and weightless string.

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