Block Colliding with a Spring - Help Appreciated

In summary, a 1.90 kg block collides with a horizontal spring with a spring constant of 237 N/m. The block compresses the spring 13.00 cm and has a coefficient of kinetic friction of 0.430. The work done by the spring in bringing the block to rest can be calculated using the equation ½ * 0.430 * 0.13 ^2. However, this does not include the loss of energy due to friction. To solve the problem, one can draw a diagram and apply kinematics equations to find the speed of the block when it hits the spring.
  • #1
Nfinley1
10
0

Homework Statement


A moving 1.90 kg block collides with a horizontal spring whose spring constant is 237 N/m.

The block compresses the spring a maximum distance of 13.00 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.430. What is the work done by the spring in bringing the block to rest?

How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?

What is the speed of the block when it hits the spring?

Homework Equations


∆K+∆U=∆E

The Attempt at a Solution


Work done by spring = ½ * 0.430 * 0.13 ^2 and many other assorted things. None of which has been correct. Simply put, I've been firing at it in the dark.Where do I need to start?
 
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  • #2
Nfinley1 said:

Homework Equations


∆K+∆U=∆E
You have left out the loss to friction.

Start by drawing a diagram and identifying the forces on the block. Then use the kinematics equations.
 
  • #3
That makes sense, I just solved it all.

Thanks for the prompt response!
 

1. How does a block colliding with a spring work?

When a block collides with a spring, the spring exerts a force on the block in the opposite direction of its displacement. This force causes the block to compress the spring, storing potential energy. When the block reaches its maximum compression, the spring will push back on the block, causing it to rebound in the opposite direction.

2. What factors affect the behavior of a block colliding with a spring?

The behavior of a block colliding with a spring is affected by the mass of the block, the stiffness of the spring, and the initial velocity of the block. A heavier block or a stiffer spring will result in a greater compression of the spring and a larger rebound. A higher initial velocity will also result in a larger rebound.

3. How is the energy conserved in a block colliding with a spring?

In a block colliding with a spring, the total energy is conserved. Initially, the block has only kinetic energy, which is converted to potential energy as the spring is compressed. At maximum compression, all of the initial kinetic energy is stored as potential energy in the compressed spring. When the spring pushes back on the block, the potential energy is converted back to kinetic energy, and the block rebounds with the same initial velocity.

4. Can the block-spring system be used to measure the spring constant?

Yes, the block-spring system can be used to measure the spring constant. By measuring the displacement of the spring and the corresponding force exerted on the block, Hooke's Law (F = kx) can be applied to calculate the spring constant. This constant represents the stiffness of the spring and can be used to predict the behavior of the system.

5. How does the surface on which the block and spring are resting affect the collision?

The surface on which the block and spring are resting can affect the collision by providing different levels of friction. A rough surface will result in more friction and therefore, a shorter distance traveled by the block before coming to rest. A smooth surface will result in less friction and a longer distance traveled. This can affect the compression and rebound of the spring, ultimately changing the behavior of the system.

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