Minimum Mass required to lift up an object from a spring

At that point, the force on m1 is equal to its weight.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. At that point, the force on m1 is equal to its weight.In summary, the question is asking for the minimum mass required for the block to lose contact with the ground when
  • #1
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Homework Statement


FInd min mass so that block loses contact with ground.
Untitled-1.jpg

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
 
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  • #2
Which mass is on the ground? Is the other mass suddenly released, or is it lowered gradually?
 
  • #3
Chestermiller said:
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
 
  • #4
I love physics said:
Spring is initially in its natural length, the mass attached to spring is on ground
Then m1 is suddenly released?
 
  • #5
Chestermiller said:
Then m1 is suddenly released?
Yes sir
 
  • #6
I love physics said:
Yes sir
Then your answer is not correct. From your equation, what is kx equal to?
 
  • #7
Chestermiller said:
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
 
  • #8
if I am wrong please correct my mistake :)
 
  • #9
I love physics said:
if I am wrong please correct my mistake :)
Like I said, from your equation, what is kx equal to?
 
  • #10
Chestermiller said:
Like I said, from your equation, what is kx equal to?
i didnt understand your question
 
  • #11
The equation ##1/2kx^2=m_1gx## is correct. What does this give you for kx?
 
  • #12
1/2kx2=m1gx
⇒kx=2m1g
⇒k2mg/k=2m1g
Therefore, M=M1
 
  • #13
I love physics said:
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##
 
  • #14
Chestermiller said:
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
 
  • #15
I love physics said:
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),
$$kx=Mg$$
 
  • #16
Chestermiller said:
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?
 
  • #17
I love physics said:
so answer should be M/2?
Sure. Can you say in words what is happening here physically?
 
  • #18
Chestermiller said:
Sure. Can you say in words what is happening here physically?
You mean more clear question statement?
 
  • #19
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
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
I love physics said:
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.
 
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FAQ: Minimum Mass required to lift up an object from a spring

What is the minimum mass required to lift up an object from a spring?

The minimum mass required to lift up an object from a spring is determined by the force exerted by the spring and the force of gravity acting on the object. This can be calculated using the equation F = kx, where F is the force, k is the spring constant, and x is the distance the spring is compressed or stretched. The minimum mass required would be the mass that creates a force equal to or greater than the force exerted by the spring.

How does the spring constant affect the minimum mass required?

The spring constant is a measure of the stiffness of a spring. A higher spring constant means the spring is stiffer and will exert a greater force. Therefore, a higher spring constant would require a lower minimum mass to lift an object from the spring compared to a spring with a lower spring constant.

Does the distance the spring is compressed or stretched affect the minimum mass required?

Yes, the distance the spring is compressed or stretched, represented by x in the equation F = kx, directly affects the minimum mass required. The greater the distance, the lower the minimum mass needed to lift an object from the spring. This is because a greater distance means a greater force is exerted by the spring.

What is the relationship between the mass of the object and the minimum mass required?

The mass of the object that needs to be lifted and the minimum mass required are directly proportional. This means that as the mass of the object increases, the minimum mass required also increases. This is because a heavier object requires a greater force to be lifted, and therefore a higher minimum mass is needed to produce that force.

Can the minimum mass required be affected by other factors?

Yes, there are other factors that can affect the minimum mass required to lift an object from a spring. These can include the strength and angle of the force applied to the object, as well as any external forces acting on the object or the spring. These factors should be taken into consideration when calculating the minimum mass required for a specific situation.

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