Solving for Distance: Spring Collision [SOLVED]

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Homework Help Overview

The problem involves a block being pushed against a spring, which is then released to determine how far the block travels along a horizontal plane before coming to a stop. The context includes concepts from mechanics, specifically energy conservation and friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the energy balance between potential energy stored in the spring and the work done against friction. There are attempts to derive the distance traveled by the block using different formulations of the energy equations.

Discussion Status

Multiple attempts to solve the problem have been shared, with participants questioning the significance of significant figures and the reference point for measuring distance. Some guidance has been offered regarding the interpretation of the distance measurement, indicating a productive direction in the discussion.

Contextual Notes

Participants note confusion regarding the reference point for measuring distance, whether from the compressed position or the original position of the block. There is also mention of the online system's rules regarding significant figures, which adds complexity to the problem-solving process.

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[SOLVED] Spring Collision

Homework Statement



A 8 kg block is pushed by an external force against a spring with spring constant 173 N/m until the spring is compressed by 2.1 m from its uncompressed length (x = 0). The block rests on a horizontal plane that has a coefficient of kinetic friction of 0.56. The acceleration of gravity is 9.8 m/s^2. Remember: The block is not attached to the spring.

After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop? Answer in units of m.



The attempt at a solution...

Ui - Wf = 1/2mv^2

(0.56)mgd = 1/2kx^2 - 1/2mv^2 then I just moved everything over to get d but I used 1/2kx^2 to get the v by setting it equal to 1/2mv^2 so it just cancels out...
 
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grouchy said:

Homework Statement



A 8 kg block is pushed by an external force against a spring with spring constant 173 N/m until the spring is compressed by 2.1 m from its uncompressed length (x = 0). The block rests on a horizontal plane that has a coefficient of kinetic friction of 0.56. The acceleration of gravity is 9.8 m/s^2. Remember: The block is not attached to the spring.

After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop? Answer in units of m.



The attempt at a solution...

Ui - Wf = 1/2mv^2

(0.56)mgd = 1/2kx^2 - 1/2mv^2 then I just moved everything over to get d but I used 1/2kx^2 to get the v by setting it equal to 1/2mv^2 so it just cancels out...
But it is given that the block comes to a stop, so v_f =?
 
attempt 2...

u = coefficient of friction)

1/2kx^2 = umg*d

d = 1/2kx^2/umg
d = 8.668616071m (wrong again)
 
grouchy said:
attempt 2...

u = coefficient of friction)

1/2kx^2 = umg*d

d = 1/2kx^2/umg
d = 8.668616071m (wrong again)
Too many significant figures! Try 8.7 meters from its released position.
 
that doesn't matter thought, the site even says "do not use" sig figs...
 
grouchy said:
that doesn't matter thought, the site even says "do not use" sig figs...
Well maybe the darn site is looking for the distance it travels from its original position before the spring was compressed, from x =0, in which case maybe it's looking for d = (8.7 -2.1) = 6.6m?
 
grouchy said:
attempt 2...

u = coefficient of friction)

1/2kx^2 = umg*d

d = 1/2kx^2/umg
d = 8.668616071m (wrong again)
Your method is fine. I would check your calculation again (I get a different 3rd digit). It shouldn't matter, but that depends on how fussy the online system is.

More important than that is where are you supposed to measure Distance from? From the compressed position (x = -2.1 m)? Or the unstretched position (x = 0)? (Based on how you described it, I'd say from the compressed position, just like you found.)
 
yup. it was -2.1 so 6.6m. thanks guys
 

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