Spring mass system and Simple harmonic motion

Click For Summary

Homework Help Overview

The problem involves a spring-mass system where a larger block is attached to a spring, and a smaller block is placed against it, compressing the spring. The scenario examines the conditions under which the smaller block begins to slip due to static friction as the system is released from rest.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the blocks, including friction and spring forces. There is an attempt to calculate the distance the second block moves before slipping, with some questioning the correctness of the approach. Others suggest drawing free body diagrams to clarify the forces involved.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's calculations and encouraging the use of free body diagrams. There are indications of differing interpretations regarding the normal force acting on the blocks.

Contextual Notes

Participants are working under the constraints of static friction and the properties of the spring, with specific values provided for mass and spring constant. The problem setup includes assumptions about the system's behavior upon release.

physicsnewb12
Messages
4
Reaction score
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?
 

Attachments

Physics news on Phys.org
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?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 18 ·
Replies
18
Views
1K
Replies
17
Views
2K
Replies
16
Views
2K
Replies
7
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
24
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
2
Views
2K