Is the Velocity Calculation Affected by Gravity or Elastic Potential Energy?

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The discussion centers on calculating the velocity of a ball at point Q after being released from a compressed spring. The original poster calculated a velocity of 2.45 m/s, while the expected answer is 3 m/s. Participants suggest using an overall potential energy approach, emphasizing the need to consider both gravitational potential energy and elastic potential energy. The calculation of elastic potential energy is confirmed as 4.5 J, and the relationship between elastic potential energy and kinetic energy is discussed, highlighting the role of gravity in the overall energy balance. The conversation underscores the importance of correctly applying energy conservation principles in solving the problem.
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Homework Statement
The spring is compressed with a force of 30 N, calculate the ball’s velocity at point Q when the compressive force is released. The mass of the ball is 300g. (Ignore air resistance).
Relevant Equations
I used mgh = 1/2mv2
Hi all!
I was attempting the following question but my answer differed from the given answer. Please help me to solve it. Thanks.

My answer = 2.45 m/s
Given answer = 3 m/s
 

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Welcome to the PF. :smile:

Can you show your work so we can check it?
 
Also, have you tried using an overall potential energy approach to solve this?
 
sacred76 said:
Problem Statement
The spring is compressed with a force of 30 N, calculate the ball’s velocity at point Q when the compressive force is released. The mass of the ball is 300g. (Ignore air resistance).
Relevant Equations
I used mgh = 1/2mv2

Hi all!
I was attempting the following question but my answer differed from the given answer. Please help me to solve it. Thanks.

My answer = 2.45 m/s
Given answer = 3 m/s

What do you think the force of ##30N## involves? Is that the force on the spring? Or is that some force applied to the ball? Are these the same?

Do you want the velocity at point Q or the velocity when the ball is released?
 
Errr...I'm lost & confused :headbang:. Please teach me how to do it. TQ

I used mgh = 1/2mv2

Working : mgh = 1/2mv2
m x 10 x 0.3 = 1/2 x m x v2
3 = 1/2v2
v2 = 6
v = 2.45 m/s
 
sacred76 said:
Errr...I'm lost & confused :headbang:. Please teach me how to do it. TQ

I used mgh = 1/2mv2

Working : mgh = 1/2mv2
m x 10 x 0.3 = 1/2 x m x v2
3 = 1/2v2
v2 = 6
v = 2.45 m/s
At point Q, the work done by the spring will be shared between the gained GPE, mgh, and the gained KE. These need not be equal.
What you have found is the KE the ball would have if dropped from point Q to the shown starting position (but with no spring there).

Can you find the energy stored in the spring before release?
 
haruspex said:
At point Q, the work done by the spring will be shared between the gained GPE, mgh, and the gained KE. These need not be equal.
What you have found is the KE the ball would have if dropped from point Q to the shown starting position (but with no spring there).

Can you find the energy stored in the spring before release?
Is it 30 N? Sorry, I'm completely lost here as I can't really comprehend the above. Could you please show me the working/equation for me to digest?
 
sacred76 said:
Is it 30 N?
That's a force, not energy.
Calculate the value of the spring constant.
 
haruspex said:
That's a force, not energy.
Calculate the value of the spring constant.

Is it using 1/2Fx?
Elastic potential energy = 1/2 x 30 N x (0.3 m)
= 4.5 J
 
  • #10
sacred76 said:
Is it using 1/2Fx?
Elastic potential energy = 1/2 x 30 N x (0.3 m)
= 4.5 J
Yes, that does it. (If you did not know the Fx/2 formula you could get there by combining F=kx and E=kx2/2.)
So relate that energy to the situation at point Q.
 
  • #11
haruspex said:
Yes, that does it. (If you did not know the Fx/2 formula you could get there by combining F=kx and E=kx2/2.)
So relate that energy to the situation at point Q.
Can I assume that it's Elastic potential energy = Kinetic energy which is 1/2Fx = 1/2mv2?
 
  • #12
sacred76 said:
Can I assume that it's Elastic potential energy = Kinetic energy which is 1/2Fx = 1/2mv2?

Don't you think that gravity might have something to say about that?
 
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