Question about spring compression

[chevyboy86]

A 12 kg block slides from rest down frictionless 35 degree incline and is stopped by spring w/ constant of 3.00x10^4 N/m. The block slides 3 m from the point of release to the point of rest against spring, how far has spring compressed?

Now I know this involves the equation Ws= 1/2kxmax2, but I am not sure where to go from there.

 

[Astronuc]

The block starts at rest and accelerates on a frictionless plane/incline. One could determined the acceleration of the block down the incline (F/m) and the velocity at the distance traveled.

Another way to do this is to apply conservation of energy. At the beginning, the energy of the block is simply gravitational potential energy (GPE). When block to the spring, it decreases in elevation, so its GPE changes, and that changes equals the increase in kinetic energy.

When the block compresses the spring, its kinetic energy is transformed into the springs potential energy.

http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html

 

[kyle8921]

Based on the information provided, we can use the equation Ws= 1/2kxmax2 to calculate the compression of the spring. Ws represents the work done by the block on the spring, k is the spring constant, and xmax is the maximum compression distance. Since the block slides 3 m from the point of release to the point of rest, we can use this distance as our xmax value.

Plugging in the values, we get Ws= 1/2(3.00x10^4 N/m)(3 m)^2 = 135,000 J. This is the amount of work done by the block on the spring. The work done by the block is equal to the potential energy stored in the spring, which can also be calculated using the equation PE= 1/2kx^2, where PE is the potential energy, k is the spring constant, and x is the compression distance. Rearranging this equation, we get x= √(2PE/k).

Substituting in the value for PE (135,000 J) and k (3.00x10^4 N/m), we get x= √(2(135,000 J)/(3.00x10^4 N/m)) = 0.58 m. Therefore, the spring has compressed by 0.58 m.

It is important to note that this calculation assumes ideal conditions, such as a frictionless surface and a perfectly vertical incline. In real-world situations, there may be other factors that could affect the compression of the spring.