Block collides with massless spring

In summary: 's up in my previous post because i was trying to compare the displacement between when the block first hit the spring and when it was fully compressed. after doing the math again and making sure the x's were lined up, it turns out the displacement and the distance past natural length is the same. so the coefficient for work done by friction is actually 1.8
  • #1
Idividebyzero
64
0
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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  • #2
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.
 
  • #3
final potential energy - initial potential energy = work?
 
  • #4
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.
 
  • #5
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
 
  • #6
bump I am still not getting it
 
  • #7
The relevant potential energy is spring PE, not gravitational PE.
 
  • #8
ah so in post 5 replace the final mgh with .5kx^2?
 
  • #9
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
 

Related to Block collides with massless spring

What is a block collides with a massless spring?

A block collides with a massless spring is a physical phenomenon where a block of mass collides with a spring that has negligible mass and undergoes a change in motion due to the interaction between the two objects.

What factors affect the collision between a block and a massless spring?

The factors that affect the collision between a block and a massless spring include the mass and velocity of the block, the stiffness and length of the spring, and the angle and surface of the collision.

What happens to the kinetic energy of the block during the collision?

During the collision between a block and a massless spring, the kinetic energy of the block is converted into potential energy stored in the spring. This energy is then released as the spring pushes the block back, causing the block to move in the opposite direction.

How does the mass of the spring affect the collision?

Since the spring is assumed to have negligible mass, the mass of the spring does not significantly affect the collision between the block and the spring. However, a heavier spring may have a different stiffness and thus affect the motion of the block differently.

What is the significance of a massless spring in this collision?

A massless spring is used in this collision because it simplifies the analysis of the motion and allows for a more accurate representation of real-world scenarios. In reality, all springs have some mass, but for the purposes of theoretical calculations, a massless spring is often assumed.

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