Finding acceleration of two masses connected by spring horizontally

AI Thread Summary
The discussion focuses on calculating the acceleration of two masses connected by a spring on a horizontal surface, considering the effects of kinetic friction. The initial approach involved applying Hook's Law and Newton's Second Law to each mass, leading to equations for their accelerations. A key realization was that treating one mass as fixed neglects the interaction between the two masses, but for the moment of release, the spring's force is determined solely by its compression. The conversation confirms that the initial accelerations can be calculated using the derived equations, assuming the spring force exceeds friction. Ultimately, the analysis emphasizes that the relative masses will influence their subsequent motion, but the initial accelerations are correctly determined.
Revivalist
Messages
1
Reaction score
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!
 
Physics news on Phys.org
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:
 
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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top