Spring mass system and Simple harmonic motion

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SUMMARY

The discussion focuses on a spring mass system involving two blocks, where a 300 gm block is attached to a spring with a spring constant of 1000 N/m, and a 100 gm block is compressed against it. The coefficient of static friction between the blocks is 0.2. The critical distance before the second block starts to slip is calculated to be 0.49 cm, derived from the equation Ff = u*N = (m_2)*g. Participants confirm the correctness of the approach and emphasize the importance of free body diagrams for clarity.

PREREQUISITES
  • Understanding of Hooke's Law and spring constant (k)
  • Knowledge of static friction and its coefficient (u)
  • Ability to apply Newton's laws of motion
  • Familiarity with free body diagrams for analyzing forces
NEXT STEPS
  • Study the dynamics of spring-mass systems in detail
  • Learn about free body diagram techniques for complex systems
  • Explore the effects of varying coefficients of friction on motion
  • Investigate energy conservation principles in oscillatory motion
USEFUL FOR

Physics students, mechanical engineers, and anyone studying dynamics and oscillatory systems will benefit from this discussion.

physicsnewb12
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Homework Statement


A block of mass 300 gm is attached to a spring with spring constant 1000 N/m. A smaller block of mass 100 gm is pushed against the first block, compressing the spring 5 cm. The coefficient of static friction between the two blocks is u = 0.2. The blocks are then released from rest.

How far from its unstretched, equilibrium position is the second block located before it just starts to slip against the first?


Homework Equations



Ff= u*N (N= normal force; u=coefficient of friction)
Fx=N=-k*x
Since x=-0.5 Fx is in positive direction or Fx=N=k*x

The Attempt at a Solution


Ff = u*N = u*k*x
Mass 2 (m_2) will start to slip when Ff = (m_2)*g.
Ff =u*k*x = (m_2)* g
or when x= m_2*g/(u*k) = 0.004905 m or 0.49 cm

Can someone tell me if i did this right?
 

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Draw a free body diagram for each block.

The normal force will NOT be -kx .
 
physicsnewb12 said:

Homework Statement


A block of mass 300 gm is attached to a spring with spring constant 1000 N/m. A smaller block of mass 100 gm is pushed against the first block, compressing the spring 5 cm. The coefficient of static friction between the two blocks is u = 0.2. The blocks are then released from rest.

How far from its unstretched, equilibrium position is the second block located before it just starts to slip against the first?


Homework Equations



Ff= u*N (N= normal force; u=coefficient of friction)
Fx=N=-k*x
Since x=-0.5 Fx is in positive direction or Fx=N=k*x

The Attempt at a Solution


Ff = u*N = u*k*x
Mass 2 (m_2) will start to slip when Ff = (m_2)*g.
Ff =u*k*x = (m_2)* g
or when x= m_2*g/(u*k) = 0.004905 m or 0.49 cm

Can someone tell me if i did this right?

For me if not going wrong:

Yup ! your concepts is right.:smile:
 
SammyS said:
Draw a free body diagram for each block.

The normal force will NOT be -kx .

What will the normal force be then?
 
physicsnewb12 said:
What will the normal force be then?

-kx is the force the spring exerts on m1. It's also the net force exerted on the combination of the two blocks if they are in contact.

What is the acceleration of the blocks when they are in contact?

Have you drawn the free body diagrams?
 

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