Solving for Distance: Spring Collision [SOLVED]

In summary, a block with a mass of 8 kg is pushed against a spring with a spring constant of 173 N/m until the spring is compressed by 2.1 m. The block is placed on a horizontal plane with a coefficient of kinetic friction of 0.56. After the external forces are removed, the block moves a distance of 6.6 m from its compressed position before coming to a stop.
  • #1
grouchy
73
0
[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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  • #2
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 =?
 
  • #3
attempt 2...

u = coefficient of friction)

1/2kx^2 = umg*d

d = 1/2kx^2/umg
d = 8.668616071m (wrong again)
 
  • #4
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.
 
  • #5
that doesn't matter thought, the site even says "do not use" sig figs...
 
  • #6
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?
 
  • #7
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.)
 
  • #8
yup. it was -2.1 so 6.6m. thanks guys
 

1. What is Spring Collision?

Spring Collision is a term used in physics to describe the interaction between two objects that collide with each other while connected by a spring. This type of collision occurs when there is a force acting on the objects, causing them to compress or stretch the spring and transfer energy to each other.

2. What factors affect the outcome of a Spring Collision?

The outcome of a Spring Collision is affected by several factors, including the mass and velocity of the objects, the stiffness and length of the spring, and the angle at which the objects collide. These factors determine the amount of energy transferred and the resulting motion of the objects after the collision.

3. How does the conservation of energy apply to Spring Collisions?

The conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In a Spring Collision, the total energy of the system before and after the collision remains constant. This means that the energy lost by one object is gained by the other, resulting in a balanced exchange.

4. Can Spring Collisions be elastic or inelastic?

Yes, Spring Collisions can be both elastic and inelastic. In an elastic collision, both objects bounce off each other and retain their original shapes and velocities. In an inelastic collision, the objects stick together and move as one, resulting in a loss of kinetic energy. The elasticity of a Spring Collision depends on the stiffness and length of the spring.

5. How is the motion of the objects affected by the spring in a Spring Collision?

The spring in a Spring Collision acts as a medium for transferring energy between the objects. As the spring compresses or stretches, it exerts a force on the objects, causing them to accelerate and change direction. The resulting motion of the objects after the collision is influenced by the properties of the spring and the initial conditions of the collision.

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