Moving two blocks connected by a spring

AI Thread Summary
The discussion revolves around a physics problem involving two blocks connected by a spring, focusing on energy conservation principles. Participants analyze kinetic and potential energy equations, attempting to calculate the final velocities of the blocks after work is done on the system. The conversation highlights the relationship between work done, kinetic energy, and potential energy, emphasizing that the total energy should equal the work done on the system. Confusion arises regarding the calculation of the center of mass velocity and the correct interpretation of energy equations. Ultimately, the conclusion is that the final energy should equal the work done, leading to a total of 8 Joules, although some calculations are debated.
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Homework Statement


m_1, v_1f
m_2, v_2f
m_2 = m_1 = m

Homework Equations



KE = .5mv^2
W = F*d
spring: F= -kx
pot energy = PE = -.5kx^2

The Attempt at a Solution



A) K + U = .5m(v_1f)^2 + .5m(v_2f)^2 - .5kx^2
.5(.1kg)(v_1f)^2 + .5(.1kg)(v_2f)^2 -.5(100N/m)(.02m)^2

B) ??
translational KE - going straight?
isnt it
.5(.1kg)(v_1f)^2 + .5(.1kg)(v_2f)^2

C) ??
V_cm...
??
am i looking for how fast the center of the spring is moving?
the center of the spring moved from .03 m to .1m

and so

?? I am lost

or do I get the average speed of the two boxes which would be

V_cm = (V_1f+V_2f)/2

or

.5(.1kg)(v_1f)^2 + .5(.1kg)(v_2f)^2 -.5(100N/m)(.02m)^2 = Wd = (5N)(.08m)
(v_1f)^2 + (v_2f)^2 = 8.4 (m^2)/(s^2)

does that equal V_cm?

V_cm = ( 8.4 (m^2)/(s^2) ) ^ (1/2) = 2.90 m/s

D) ??

vibrational?
 

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(A)
How much work has been done on this system?
 
Wd = (5N)(.08m)
 
Okay, so the energy system has increased by that amount. That helps with (A), getting the total energy K+U. Do you see why?
 
what are you trying to say?

the initial energy = 0
becuz the objects are at rest and spring is at its natural lenght

so therefore

the final energy is K + U which is equal to Wd

that?

and so

.5m(v_1f)^2+.5m(v_2f)^2-.5kx^2 = (100N/m)(.08m)
KE m1 KE m2 PEspr Wd

so my answer is??

8 Joules
 
soupastupid said:
what are you trying to say?

the initial energy = 0
becuz the objects are at rest and spring is at its natural lenght
Yes, correct.

so therefore

the final energy is K + U which is equal to Wd

that?

Almost. K+U is equal to the work done, W=Fd:

Wd = (5N)(.08m)
Well, this is really W or Fd, but not Wd. At any rate, multiplying (5N) times (0.08m) will give you the answer to (A).

and so

.5m(v_1f)^2+.5m(v_2f)^2-.5kx^2 = (100N/m)(.08m)
KE m1 KE m2 PEspr Wd

so my answer is??

8 Joules
No, see above comments.
 
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