Spring mass system and Simple harmonic motion

In summary, a smaller block of mass 100 gm is pushed against a block of mass 300 gm attached to a spring with spring constant 1000 N/m, compressing the spring 5 cm. The coefficient of static friction between the two blocks is u = 0.2. When released from rest, the second block will start to slip against the first when it is located 0.49 cm from its unstretched, equilibrium position. The normal force between the blocks will not be -kx, but rather the force the spring exerts on the first block, and the acceleration of the blocks in contact will depend on the individual free body diagrams.
  • #1
physicsnewb12
4
0

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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  • #2
Draw a free body diagram for each block.

The normal force will NOT be -kx .
 
  • #3
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:
 
  • #4
SammyS said:
Draw a free body diagram for each block.

The normal force will NOT be -kx .

What will the normal force be then?
 
  • #5
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?
 

1. What is a spring-mass system?

A spring-mass system is a physical system consisting of a mass attached to a spring. When the mass is displaced from its equilibrium position, the spring exerts a restoring force that causes the mass to oscillate back and forth around its equilibrium position. This system is commonly used in physics and engineering to model simple harmonic motion.

2. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction of the displacement. This results in a sinusoidal motion, with the mass oscillating back and forth around its equilibrium position with a constant amplitude and frequency.

3. How is the period of a spring-mass system calculated?

The period of a spring-mass system can be calculated using the formula T = 2π√(m/k), where T is the period in seconds, m is the mass in kilograms, and k is the spring constant in Newtons per meter. This formula assumes that there is no damping or external forces acting on the system.

4. What factors affect the frequency of a spring-mass system?

The frequency of a spring-mass system is affected by the mass of the object, the spring constant, and the amplitude of the oscillation. Increasing the mass or decreasing the spring constant will decrease the frequency, while increasing the amplitude will increase the frequency. Additionally, the presence of damping or external forces can also affect the frequency.

5. How is energy conserved in a spring-mass system?

In a spring-mass system, the total mechanical energy (the sum of kinetic and potential energy) is conserved. As the mass oscillates, it alternates between kinetic energy (when it is moving) and potential energy (when it is at the highest or lowest point of its oscillation). This energy exchange allows the system to continue oscillating without losing energy, as long as there is no external friction or damping present.

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