# Moving two blocks connected by a spring

1. Nov 30, 2008

### soupastupid

1. The problem statement, all variables and given/known data
m_1, v_1f
m_2, v_2f
m_2 = m_1 = m

2. Relevant equations

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

3. 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

?? im 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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2. Nov 30, 2008

### Redbelly98

Staff Emeritus
(A)
How much work has been done on this system?

3. Nov 30, 2008

### soupastupid

Wd = (5N)(.08m)

4. Dec 1, 2008

### Redbelly98

Staff Emeritus
Okay, so the energy system has increased by that amount. That helps with (A), getting the total energy K+U. Do you see why?

5. Dec 1, 2008

### soupastupid

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

8 Joules

6. Dec 1, 2008

### Redbelly98

Staff Emeritus
Yes, correct.

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

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).