Block collides with massless spring

AI Thread Summary
A 2 kg block collides with a massless spring (spring constant 109 N/m) at a speed of 1.5 m/s, compressing the spring to a maximum distance of 12.1911 cm on a frictionless surface. The discussion revolves around calculating the kinetic coefficient of friction between the block and the floor, with initial attempts yielding incorrect results. Participants clarify that the work done by friction should equal the difference between initial and final energy states, emphasizing the use of spring potential energy instead of gravitational potential energy. A misunderstanding about the displacement distance leads to confusion, but ultimately, the correct approach involves recognizing that the distance the spring compresses is not the same as the distance over which friction acts. The final consensus reveals that a calculation error occurred due to not squaring the compression distance in the energy equations.
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
 
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  • #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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