# Spring mass system and Simple harmonic motion

## 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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SammyS
Staff Emeritus
Homework Helper
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Draw a free body diagram for each block.

The normal force will NOT be -kx .

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

The normal force will NOT be -kx .

What will the normal force be then?

SammyS
Staff Emeritus