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## Homework Statement

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.

http://imgur.com/5JF0F

**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:

x_final=x_initial+v_initial*t+1/2*a*t^2

.318=0+0.75*t+1/2*(9.8*sin(20 degrees))*t^2

t=0.26951 seconds

now that we have time let's get final velocity.

v_final=v_initial+a*t

v_final=0.75+9.8*sin(20 degrees)*0.26951

v_final=1.653 m/s

Now let's get the kinetic energy of the block when it hits the spring.

Kinetic=m*v^2/2

2.37*1.653^2/2=3.238J

Now let's get the potential energy on the spring

U_spring=1/2*k*(x_final^2-x_initial^2)

U_spring=1/2*495*(x^2-0)

495/2*x^2 J

now let's solve

3.238=495/2*x^2

x=0.114m. This is incorrect.

Where have I gone wrong?

Thanks!