# Homework Help: Minimum Mass required to lift up an object from a spring

1. Aug 8, 2017

### I love physics

1. The problem statement, all variables and given/known data
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)
2. Relevant equations
Equations of conservation of energy

3. 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 im right or not

2. Aug 8, 2017

### Staff: Mentor

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

3. Aug 8, 2017

### I love physics

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

4. Aug 8, 2017

### Staff: Mentor

Then m1 is suddenly released?

5. Aug 8, 2017

Yes sir

6. Aug 8, 2017

7. Aug 8, 2017

### I love physics

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

8. Aug 8, 2017

### I love physics

if im wrong please correct my mistake :)

9. Aug 8, 2017

### Staff: Mentor

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

10. Aug 8, 2017

### I love physics

11. Aug 8, 2017

### Staff: Mentor

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

12. Aug 8, 2017

### I love physics

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

13. Aug 8, 2017

### Staff: Mentor

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

14. Aug 8, 2017

### I love physics

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

15. Aug 8, 2017

### Staff: Mentor

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),
$$kx=Mg$$

16. Aug 8, 2017

### I love physics

17. Aug 8, 2017

### Staff: Mentor

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

18. Aug 8, 2017

### I love physics

You mean more clear question statement?

19. Aug 8, 2017

### Staff: Mentor

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?

20. Aug 8, 2017

### 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

21. Aug 8, 2017

### Staff: Mentor

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.