Minimum Mass required to lift up an object from a spring

[I love physics]

Homework Statement

FInd min mass so that block loses contact with ground.

Spring is initially in its natural length, the mass attached to spring is on ground(M)

Homework Equations

Equations of conservation of energy

The Attempt at a Solution

Min compression = 2mg/k to lift up object, Normal rxn=0
1/2kx²=m₁gx
⇒ m=m₁

Can someone tell me if I am right or not

 

[Chestermiller]

Which mass is on the ground? Is the other mass suddenly released, or is it lowered gradually?

 

[I love physics]

Which mass is on the ground? Is the other mass suddenly released, or is it lowered gradually?

Spring is initially in its natural length, the mass attached to spring is on ground

 

[Chestermiller]

Spring is initially in its natural length, the mass attached to spring is on ground

Then m1 is suddenly released?

 

[I love physics]

Then m1 is suddenly released?

Yes sir

 

[Chestermiller]

Yes sir

Then your answer is not correct. From your equation, what is kx equal to?

 

[I love physics]

Then your answer is not correct. From your equation, what is kx equal to?

k is spring constant and x is displacement from its natural length
1/2kx^2 is potential energy stored in spring which is equal to gravitational potential which is M1gx

 

[I love physics]

if I am wrong please correct my mistake :)

 

[Chestermiller]

if I am wrong please correct my mistake :)

Like I said, from your equation, what is kx equal to?

 

[I love physics]

Like I said, from your equation, what is kx equal to?

i didnt understand your question

 

[Chestermiller]

The equation $1/2kx^2=m_1gx$ is correct. What does this give you for kx?

 

[I love physics]

1/2kx²=m₁gx
⇒kx=2m₁g
⇒k2mg/k=2m1g
Therefore, M=M1

 

[Chestermiller]

1/2kx²=m₁gx
⇒kx=2m₁g
⇒k2mg/k=2m1g
Therefore, M=M1

That's not what I get. I get $$kx=Mg=2m_1g$$
So, $M=2m_1$

 

[I love physics]

That's not what I get. I get $$kx=Mg=2m_1g$$
So, $M=2m_1$

The compression in spring should be 2mg/k in order to lift it up, therefore i replaced x by 2mg/k

 

[Chestermiller]

The compression in spring should be 2mg/k in order to lift it up, therefore i replaced x by 2mg/k

Force balance on M: $$N+kx=Mg$$where N is the upward normal force exerted by the table on M. When N = 0 (M loses just loses contact with the table),
$$kx=Mg$$

 

[I love physics]

Force balance on M: $$N+kx=Mg$$where N is the upward normal force exerted by the table on M. When N = 0 (M loses just loses contact with the table),
$$kx=Mg$$

so answer should be M/2?

 

[Chestermiller]

so answer should be M/2?

Sure. Can you say in words what is happening here physically?

 

[I love physics]

Sure. Can you say in words what is happening here physically?

You mean more clear question statement?

 

[Chestermiller]

No. I mean mechanistically the reason that, in order for M to lose contact, $m_1\geq M/2$. Certainly, if m1 were lowered slowly, the condition for loss of contact would be $m_1\geq M$. So why, in the case where $m_1$ is released quickly, can $m_1$ weigh less and still cause M to lose contact even if it is only half of M? What is the motion of $m_1$ like?

 

[I love physics]

If it is released from rest it will produce potential energy energy in the spring, which will further be transferred to our original block M
and if it is lowered slowly change in potential energy will be negligible

 

[Chestermiller]

If it is released from rest it will produce potential energy energy in the spring, which will further be transferred to our original block M
and if it is lowered slowly change in potential energy will be negligible

Here's my spin on this. After you release m1 quickly, it begins falling, but then overshoots the equilibrium displacement by a factor of 2. So the tension in the cord and spring when m1 reaches its lowest point will be 2x the value at the equilibrium displacement. If, on the other hand, m1 is lowered slowly (say by hand), it will stop moving downward when it is at 1x the equilibrium displacement.