Conservation of Energy on an Inclined Plane with Spring

[bcca]

Hello. Can someone help me with this problem please?

An inclined plane of angle theta = 20.0 has a spring of force constant k=500 N/M fastened securely at the bottom so that the spring is parallel to the surface as shown in figure P7.63. A block of mass m = 2.50 kg is placed on the plane at a distance d = .300 m from the spring. From this position, the block is projected downward toward the spring with speed v = .750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?

In short:
Angle: theta = 20.0
Force constant: k = 500 N/M
Mass: m = 2.50 kg
Distance: d = .300 m
Speed: v = .750 m/s
Compression: x = ?

The answer is .131 m. I’m supposed to use conservation of energy. I’ve tried to solve it by choosing my reference level to be at the top of the fully compressed spring, so the height of the block fallen would be d*sin(theta)+x*sin(theta). I figure K+Ug=Us and I can solve for x with a quadratic, but it doesn't work. Sorry I don’t have a diagram :/.

Attempt:
E = E’
K + Ug = Us
1/2mv² + mgd*sin(theta) + mgx*sin(theta) = 1/2kx²
1/2kx² – 1/2mv² – mgd*sin(theta) – mgx*sin(theta) = 0
250x² – 8.379x – 1.8107 = 0
(solving the quadratic) x = .1035 m

Thanks in advance!

 

[rl.bhat]

Hi bcca, welcome to PF.

1/2mv2 + mgd*sin(theta) + mgx*sin(theta)= 1/2kx2

1/2mv2 + mgd*sin(theta)

check this calculation.

 

[bcca]

Oh, thank you so much! That was a very persistent calculation error :). I kept subtracting them on accident.