Solve Halfway Measures: Find KE at 7cm

[anyone1979]

[SOLVED] Halfway measures

Am I doing this right?

You stretch the spring of a certain mass-spring system 14.0 cm from its relaxed state; this requires 21.2J of work. You then release the system. What is its Kinetic energy when the stretch of the spring is 7 cm?

W = F*s
21.2 = F(.14)
F = 151.4N

W = 151.4(.07)
W = 10.6J
W = change in KE
KE = 10.6J

 

[Doc Al]

No, you cannot assume that the spring force is constant, which is what you are doing when you set the work equal to F*s. Hint: What's the energy stored in a stretched spring? (Look it up or derive it.)

 

[anyone1979]

Is this close?

W = 1/2(kx^2)
21.2 = (1/2)k(.14^2)
21.2(2) = k(.14^2)
k = 216.3J

W = (1/2)(216.3)(.07^2)
W = .053J
KE = .053J

 

[Doc Al]

Is this close?

W = 1/2(kx^2)
21.2 = (1/2)k(.14^2)
21.2(2) = k(.14^2)
k = 216.3J

Much better. (You are close.)

You have a typo (or arithmetic error) in that last step: k = 2163 J.

Realize that this is the total energy stored in the system. After you release it, some of that initial spring potential energy will be transformed to KE. But at all times, Total energy = KE + PE.

 

[anyone1979]

Thanks for replying.
I do not see the error though. I changed it from cm to meters.

21.2 = (1/2)k(0.14^2)
21.2(2) = k(0.196)
((21.2)(2))/(0.196) = k am I calculating wrong?

U1 = (1/2)k(x1^2)
U2 = (1/2)k(x2^2)
KE1 = (1/2)mv1^2
KE2 = (1/2)mv2^2

KE2 = KE1 + U1 - U2

Is that the right Idea?

 

[Doc Al]

Thanks for replying.
I do not see the error though. I changed it from cm to meters.

21.2 = (1/2)k(0.14^2)
21.2(2) = k(0.196)
((21.2)(2))/(0.196) = k am I calculating wrong?

0.14^2 = 0.0196 (not 0.196)

U1 = (1/2)k(x1^2)
U2 = (1/2)k(x2^2)
KE1 = (1/2)mv1^2
KE2 = (1/2)mv2^2

KE2 = KE1 + U1 - U2

Is that the right Idea?

Yes, exactly the right idea.

 

[anyone1979]

Thank you so much.
I wrote the equation down wrong.