Hello, folks. I know this same problem has been answered before here but the previous responses are just not doing it for me, sorry!
An inclined plane of angle θ = 20.0° has a spring of force constant k = 495 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.37 kg is placed on the plane at a distance d = 0.318 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?
Here is an image of such a setup.
2. The attempt at a solution
So basically I want to find the displacement in the spring given these values. What I do first is get the velocity at which the box hits the spring so I can get it's kinetic energy at that point.
It starts at .750m/s but it's acceleration due to gravity is 9.8*sin(20 degrees) since it's on an incline and it has to travel .318m to hit the spring so I've got:
now that we have time let's get final velocity.
Now let's get the kinetic energy of the block when it hits the spring.
Now let's get the potential energy on the spring
now let's solve
x=0.114m. This is incorrect.
Where have I gone wrong?