Moving two blocks connected by a spring

In summary, the conversation discusses finding the final energy of a system involving two objects and a spring, given the masses and velocities of the objects and the spring constant. The final energy is equal to the work done on the system, which can be calculated using the formula W=Fd. The initial energy of the system is 0, as the objects are at rest and the spring is at its natural length. The final energy can be calculated by adding the kinetic energy of each object and the potential energy of the spring. The final answer for the final energy is 8 joules.
  • #1
soupastupid
34
0

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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  • #2
(A)
How much work has been done on this system?
 
  • #3
Wd = (5N)(.08m)
 
  • #4
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
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
 
  • #6
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.
 

FAQ: Moving two blocks connected by a spring

How does the spring affect the movement of the two blocks?

The spring acts as a connecting force between the two blocks, allowing them to move together in a coordinated manner. As one block moves, it pulls or pushes on the spring, which then exerts a force on the other block to move in the same direction.

What factors influence the motion of the blocks connected by a spring?

The motion of the blocks is influenced by several factors, including the stiffness of the spring, the mass of the blocks, and the initial displacement of the blocks from their equilibrium position.

How does the amplitude of the motion change over time?

The amplitude of the motion decreases over time due to the damping effect of the spring. This means that the blocks will gradually come to rest at their equilibrium position.

Can the spring break or wear out over time?

Yes, the spring can break or wear out over time due to repeated stretching and compressing. This can affect the overall movement of the blocks and may eventually cause the spring to fail completely.

Is it possible for the blocks to move at different frequencies?

Yes, if the two blocks have different masses or if one block is at a different initial displacement, they may move at different frequencies. This is known as a coupled oscillation and can result in complex motion patterns.

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