Block collides with massless spring

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Homework Help Overview

The problem involves a 2 kg block colliding with a massless spring with a spring constant of 109 N/m. The block's speed at the moment of collision is 1.5 m/s, and the scenario assumes a frictionless surface. The discussion centers around determining how far the spring compresses and finding the coefficient of friction between the block and the floor.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore energy conservation principles, discussing the relationship between kinetic energy, potential energy in the spring, and work done by friction.
  • Some participants question the assumptions made regarding energy loss due to friction and the definitions of potential energy involved.
  • There is a focus on the correct interpretation of the distances involved in the calculations related to spring compression and friction.

Discussion Status

The discussion is ongoing, with participants providing guidance on the correct application of energy conservation. There are multiple interpretations being explored regarding the relationship between the distances involved in the problem, and some participants express confusion about their calculations.

Contextual Notes

Participants note that the problem may involve assumptions about energy loss and the definitions of potential energy, particularly distinguishing between spring potential energy and gravitational potential energy.

Idividebyzero
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1. A 2 kg block collides with a massless spring of spring constant 109 N/m attached to a wall. The speed of the block was observed to be 1.5 m/s at the moment of collision.
The acceleration of gravity is 9.8 m/s2 How far does the spring compress if the sur-
face on which the mass moves is frictionless? The maximum distance to which the
spring was compressed was observed to be 12.1911 cm. What is the kinetic coefficient of friction between the block and the floor?




2.W=1/2kx^2



3. i have gotten the first part of the problem correct.

W=1/2kx^2
1/2mv^2-0=1/2kx^2
20.3=x

for the second part I am stumped. this is what I've tried

1/2kx^2=uk(mg)d
solving for u i get 0.94 as the coefficient but is incorrect
 
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You assumed that the final potential energy is equal to the energy lost by friction, which is incorrect. Instead, the energy lost by friction should be equal to the difference between the initial energy and the final energy.
 
final potential energy - initial potential energy = work?
 
Final total energy - initial total energy = work done by friction, because of conservation of energy. The final total energy is purely potential; the initial total energy is purely kinetic.
 
i think i got it but to be sure:

ME final -ME initial

(delta KE + delta U) final - (delta KE + delta U) initial = work
(0 + mgh) final - (1/2mv^2 + 0) initial = work

where work is W= F*D
F= (mu)(mg)*D
 
bump I am still not getting it
 
The relevant potential energy is spring PE, not gravitational PE.
 
ah so in post 5 replace the final mgh with .5kx^2?
 
Idividebyzero said:
ah so in post 5 replace the final mgh with .5kx^2?
Yes.
 
  • #10
haha that makes sense

(0 + .5kx^2) final - (1/2mv^2 + 0) initial = mu(mg)d

the X in the final energy is the distance the spring was compressed was 0.121911 m. is this the same as the discplacement for the work done by friction? i first thought that it was the same, the initial position where the block first hit the spring to the point where the spring "takes" all the blocks kinetic energy and stores it as potential energy. so the displacement and the distance past natural length is the same?

doing the math that way gives me a coefficient greater than 1.0, i get 1.8 which seems too high. this tells me x and d are not the same
 
Last edited:
  • #11
Idividebyzero said:
doing the math that way gives me a coefficient greater than 1.0, i get 1.8 which seems too high. this tells me x and d are not the same
Redo your calculation. There's only one distance mentioned in this problem.
 
  • #12
Doc Al said:
Redo your calculation. There's only one distance mentioned in this problem.

ideasrule said:
Final total energy - initial total energy = work done by friction, because of conservation of energy. The final total energy is purely potential; the initial total energy is purely kinetic.


plus 1 invisisble rep guys thanks i finally got it. its actually kinda funny i didnt sqaure the x
 

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