Spring mass system and Simple harmonic motion

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A block of mass 300 gm is attached to a spring with a spring constant of 1000 N/m, while a smaller block of 100 gm compresses the spring by 5 cm. The coefficient of static friction between the blocks is 0.2, and the problem involves determining how far the second block moves from its equilibrium position before slipping occurs. The frictional force is calculated using Ff = u*N, and the slipping condition is established when Ff equals the weight of the second block. The calculations suggest that the second block will start to slip at approximately 0.49 cm from the unstretched position of the spring. The discussion emphasizes the importance of free body diagrams to analyze the forces acting on both blocks.
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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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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