Find the initial speed of a block before it hits a spring

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SUMMARY

The discussion centers on calculating the initial speed of a block before it impacts a spring, utilizing the work-energy principle. Participants applied the equation W = ΔKE, where W represents work done by the spring, and ΔKE is the change in kinetic energy. The initial calculations yielded speeds of 3.22 m/s and 3.17 m/s, but the correct initial speed, after considering gravitational potential energy, was determined to be 2.8 m/s. The importance of accounting for both kinetic and potential energy in energy conservation equations was emphasized.

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isukatphysics69
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Homework Statement


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Homework Equations


W =ΔKE

The Attempt at a Solution


found work done by spring
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-1.3 =ΔKE
-1.3 = KEFINAL - KEINITIAL
-1.3 = .5(0.25)(VFINAL2) - .5(0.25)(VINITIAL2)

Since v final is 0 since it stops momenteraly at the bottom of the spring
-1.3 = - .5(0.25)(VINITIAL2)
sqrt((-1.3/-.125)) = VINITIAL
VINITIAL = 3.22m/s
 

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isukatphysics69 said:

Homework Statement


View attachment 224911
View attachment 224915

Homework Equations


W =ΔKE

The Attempt at a Solution


found work done by spring
View attachment 224913-1.3 =ΔKE
-1.3 = KEFINAL - KEINITIAL
-1.3 = .5(0.25)(VFINAL2) - .5(0.25)(VINITIAL2)

Since v final is 0 since it stops momenteraly at the bottom of the spring
-1.3 = - .5(0.25)(VINITIAL2)
sqrt((-1.3/-.125)) = VINITIAL
VINITIAL = 3.22m/s
I assume you want to know why the answer to question 3 is incorrect.
I see you have answered the work question with a precision of 2 significant digits. Then you used it to calculate speed at a precision of 3 significant digits.
 
I don't think that is the problem here, i think i am doing something wrong with the calculation logic here
 
-1.26 =ΔKE
-1.26 = KEFINAL - KEINITIAL
-1.26 = .5(0.25)(VFINAL2) - .5(0.25)(VINITIAL2)

Since v final is 0 since it stops momenteraly at the bottom of the spring
-1.26 = - .5(0.25)(VINITIAL2)
sqrt((-1.26/-.125)) = VINITIAL
VINITIAL = 3.17m/susing the unrounded answer, still incorrect
 
So i was using the value of work for the spring on the block to find the initial velocity of the block, i am not sure if that is correct
 
isukatphysics69 said:
-1.26 =ΔKE
-1.26 = KEFINAL - KEINITIAL
-1.26 = .5(0.25)(VFINAL2) - .5(0.25)(VINITIAL2)

Since v final is 0 since it stops momenteraly at the bottom of the spring
-1.26 = - .5(0.25)(VINITIAL2)
sqrt((-1.26/-.125)) = VINITIAL
VINITIAL = 3.17m/susing the unrounded answer, still incorrect
Yes, but you have given too many significant digits in the answer.
 
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isukatphysics69 said:
So i was using the value of work for the spring on the block to find the initial velocity of the block, i am not sure if that is correct
It looks right to me. I got the same answer independently.
 
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tnich said:
It looks right to me. I got the same answer independently.
i just tried 3.2m/s incorrect
 
isukatphysics69 said:
i just tried 3.2m/s incorrect
I think we forgot about gravitational potential energy.
 
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  • #10
tnich said:
I think we forgot about gravitational potential energy.
my book says ΔKE = KFINAL - KINITIAL = WA+WG
since energy is conserved isn't WA = -WG
 
  • #11
isukatphysics69 said:
my book says ΔKE = KFINAL - KINITIAL = WA+WG
since energy is conserved isn't WA = -WG
I don't know what WA and WG represent.
Total energy (kinetic + potential) is conserved. There is definitely a change in gravitational potential energy between the point at which the block hits the spring and the point at which it bottoms out.
 
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  • #12
tnich said:
I don't know what WA and WG represent.
Total energy (kinetic + potential) is conserved. There is definitely a change in gravitational potential energy between the point at which the block hits the spring and the point at which it bottoms out.
youre right
 
  • #13
i remember from class last week prof was talking about potential energy.
 
  • #14
tnich said:
I don't know what WA and WG represent.
Total energy (kinetic + potential) is conserved. There is definitely a change in gravitational potential energy between the point at which the block hits the spring and the point at which it bottoms out.
So you need to add up the potential energy (due to gravity and the spring) and kinetic energy. Do this at both the point where the block hits the spring and where it bottoms out.
Also, think about the appropriate sign for the potential energy of the spring in this equation.
 
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  • #15
So then i think the gravity force has to be integrated from initial to bottom
 
  • #16
wait i already have the work that gravity did while spring is compressed
 
  • #17
it was 0.29J
 
  • #18
omg i am an idiot
 
  • #19
isukatphysics69 said:
it was 0.29J
OK, you have all of the pieces. Now you need to write your conservation of energy equation.
 
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  • #20
ok the answer is 2.8 m/s
 
  • #21
isukatphysics69 said:
ok the answer is 2.8 m/s
Yes. That's what I got, too.
 
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  • #22
thank you i completely forgot about potential energy
 

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