Energy conservation with tension

In summary, the conversation discusses the calculation of the velocity of an object after it has been released from a spring in different positions. The potential energy of the spring is calculated using Hooke's law and the equation (1/2)kx^2, where x is the amount by which the spring is stretched from its unstretched length. The final equation for calculating the velocity is mg(y2-y1) + 1/2k(s2-s1)^2 = 1/2mv2^2, with x representing the length of the spring in different positions. The correct answer is determined to be 7.02 m/s.
  • #1
nysnacc
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3

Homework Statement


upload_2016-10-6_10-13-44.png


Homework Equations


Conservative of energy

mg(y2-y1) +1/2 k (s2-s02) = 1/2 mv12 +1/2 mv22

The Attempt at a Solution


v1 = 0 at rest
y2 = 0 bottom

What I got is v2 = 8.20 m/s but not correct,
I don't know how I can take into account the tension..
Fspring = -ks = -4000 N/m * 0.2 m = 800 N at position 2, isn't it? why it says 500 N?
 
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  • #2
In the formulas F = -ks and U = (1/2)ks2, does s represent the length of the spring or does it represent how much the spring is stretched?
 
  • #3
U is for how much the string stretches.
And the force is also, but it does not say whether the string is unstretch when it is in position 2
 
  • #4
If it is of lowest tension in pos 2, then in pos 1, the tension and U will be higher than it usually is.
 
  • #5
nysnacc said:
U is for how much the string stretches.
And the force is also, but it does not say whether the string is unstretch when it is in position 2
If the spring were unstretched in position 2 then it would exert no tension force in position 2.
 
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  • #6
TSny said:
If the spring were unstretched in position 2 then it would exert no tension force in position 2.
you're right! and would it be 0.125 m for the length when it is unstretched? because I found that T= ks => 500 N= 4000 N/m * s
SO s = 0.125 (the string is stretched for 0.125 m in pos 2)
 
  • #7
Oh no, should be 0.2 - 0.125 as s0 (unstretch)
 
  • #8
Just to be clear what you are saying, what are your answers for the following
(1) What is the natural (unstretched) length of the spring?
(2) How much is the spring stretched beyond its natural length in postion 2?
 
  • #9
(1) 0.075 m
(2) 0.125 m

and so in pos 1, the length stretched will be length of the spring - 0.075 m
Correct ? Thanks
 
  • #10
nysnacc said:
(1) 0.075 m
(2) 0.125 m

and so in pos 1, the length stretched will be length of the spring - 0.075 m
Correct ? Thanks
Yes, that's right.
 
  • #11
Would the equation be:

mgy1 + 1/2 mv12 + 1/2k(s12-s02) = mgy2 + 1/2 mv22+ 1/2k(s22-s02)

For the bold part, will it be zero, because it is perpendicular to the motion?
 
  • #12
nysnacc said:
Would the equation be:

mgy1 + 1/2 mv12 + 1/2k(s12-s02) = mgy2 + 1/2 mv22+ 1/2k(s22-s02)

For the bold part, will it be zero, because it is perpendicular to the motion?
The expressions for the potential energy of the spring at positions 1 and 2 are not correct. When the spring is in position 2, the spring is stretched a certain amount from its natural length. So, the potential energy of the spring at position 2 is not zero.

The fact that the spring is perpendicular to the motion at point 2 does not affect the potential energy at that point. The force of the spring is perpendicular to the motion at position 2, which means that at that instant the spring force is not doing any work. So, the potential energy of the spring is not changing at that one instant, but the spring does have nonzero potential energy at that instant.
 
  • #13
mgy1 + 1/2 mv12 + 1/2k(s12-s02) = mgy2 + 1/2 mv22+ 1/2k(s22-s02)

So I can assume the height y2 be zero, and y1 be 0.25 m
Given v1 is 0, v2 is what we are finding.

But do we need to know the spring at unstretch state? Cuz we need the change of energy?
ΔUweight + ΔUspring = ΔKE
mg(y2 -y1) + 1/2 k (s22-s12) = 1/2 m (v22-v12) ??
Loss in Uweight + Loss in Uspring = Gain in KE
 
  • #14
You are not calculating the potential energy of the spring correctly. Hooke's law is F = -kx and the potential energy of the spring is (1/2)kx2 where x is the amount by which the spring is stretched from its unstretched length. If s0 is the unstretched length of the spring and if s is the length of the spring when it is stretched, how would you express x in terms of s and s0? Then how would you express the potential energy in terms of s and s0,
 
  • #15
nysnacc said:
TSny said:
You are not calculating the potential energy of the spring correctly. Hooke's law is F = -kx and the potential energy of the spring is (1/2)kx2 where x is the amount by which the spring is stretched from its unstretched length. If s0 is the unstretched length of the spring and if s is the length of the spring when it is stretched, how would you express x in terms of s and s0? Then how would you express the potential energy in terms of s and s0,

So I can assume the height y2 be zero, and y1 be 0.25 m
Given v1 is 0, v2 is what we are finding.

But do we need to know the spring at unstretch state? Cuz we need the change of energy?
ΔUweight + ΔUspring = ΔKE
mg(y2 -y1) + 1/2 k (s22-s12) = 1/2 m (v22-v12) ??
Loss in Uweight + Loss in Uspring = Gain in KE
mgy1 + 1/2 mv12 + 1/2k(x12) = mgy2 + 1/2 mv22+ 1/2k(x22)
where x1 is length of spring in pos 1 - unstretch
x2 is length of spring in pos 2 - unstretch?
 
  • #16
So x1 = 0.245 and x2 = 0.125?
where the equation now be
mg(y2 - y1)+ 1/2k(x22-x12) = 1/2 mv22

(4)(-9.81)(0-0.25) + 1/2 (-4000) (0.1252-0.2452) = 1/2 (4) v2

9.81 +2000(0.444) = 2 v2
v = 7.02 m/s

Correct?
 
  • #17
nysnacc said:
So x1 = 0.245 and x2 = 0.125?
where the equation now be
mg(y2 - y1)+ 1/2k(x22-x12) = 1/2 mv22

(4)(-9.81)(0-0.25) + 1/2 (-4000) (0.1252-0.2452) = 1/2 (4) v2

9.81 +2000(0.444) = 2 v2
v = 7.02 m/s

Correct?
You got the right answer, but there are sign errors in your first equation above.
g and k are positive numbers.

You wrote the energy equation correctly in post #15, but you did not rearrange it correctly when you solved for 1/2 mv22.
 
  • #18
So on the left hand side, 1 - 2 then on the right hand side 1/2 m (v2^2 - v1^2), this case 2 -1 ??
 
  • #19
nysnacc said:
So on the left hand side, 1 - 2 then on the right hand side 1/2 m (v2^2 - v1^2), this case 2 -1 ??
Yes
 
  • #20
Great thanks!
I think what I was off is the calculation of x1 and x2...
I thought it was 1/2 k (S12 - S02) + the other energy = 1/2 k (S22 - S02) + other energy

where S0 is the un-stretched distance
 
  • #21
nysnacc said:
Great thanks!
I think what I was off is the calculation of x1 and x2...
I thought it was 1/2 k (S12 - S02) + the other energy = 1/2 k (S22 - S02) + other energy

where S0 is the un-stretched distance
Right. It should be 1/2 k (S1 - S0)2 rather than 1/2 k (S12 - S02) .
 
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  • #22
I see so I should calculate each "stretched distance" in pos 1 and pos 2... then find the difference between them using the equation you jjust provided.
 
  • #23
nysnacc said:
I see so I should calculate each "stretched distance" in pos 1 and pos 2... then find the difference between them using the equation you jjust provided.
I think so, if I'm understanding your statement. The key point is that U = (1/2)kx2 where x is the amount the spring has been stretched from its natural length. So, you just need to do whatever is necessary to find x. If you know the natural length (s0) of the spring, then x = s - s0 where s is the total length of the spring.
 
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1. What is tension and how does it relate to energy conservation?

Tension is the force that is exerted on an object when it is stretched or pulled. In the context of energy conservation, tension can play a role in transferring energy from one object to another, such as in the case of a spring or a pulley system.

2. How can tension be used to conserve energy?

Tension can be used to conserve energy by converting potential energy into kinetic energy. This can be seen in systems where tension is used to stretch a spring or lift an object, which then releases the stored energy to perform work.

3. What types of energy can be conserved with tension?

Tension can be used to conserve various forms of energy, such as mechanical, electrical, and gravitational potential energy. This is because tension can be applied to different types of systems, such as springs, ropes, and cables, to store and transfer energy.

4. How does tension affect the efficiency of energy conservation?

Tension can have a significant impact on the efficiency of energy conservation. If there is too much tension in a system, it can lead to losses in energy due to friction and heat. On the other hand, the right amount of tension can maximize the transfer of energy and improve efficiency.

5. Are there any limitations to using tension for energy conservation?

While tension can be a useful tool for energy conservation, there are some limitations to consider. For example, tension can only be applied in certain types of systems and may not be effective in all situations. Additionally, tension can also lead to potential safety hazards if not properly managed.

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