(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An inclined plane of angle θ = 20.0° has a spring of force constant k = 455 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 = 2.71 kg is placed on the plane at a distance d = 0.330 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?

2. Relevant equations

dE=dk+du=0

Kf-Ki+Dfsp-Disp+Dfg-Dig=0

Ki=1/2mv^2

Dfsp=1/2kx^2

Dig= mg[h + x(sin20.0)]

3. The attempt at a solution

Kf,Disp, and Dfg all Equal zero

so by substitution:

-1/2mv^2+1/2kx^2-mg[h + x(sin20.0)]=0

-(2.71)(.75^2)/2+455/2x^2-(2.71*9.8)[0.33+xsin(20)]=0

which simplifies to

227.5x^2-9.084x-9.52=0

My answer is .226, which is the incorrect answer. What am I doing wrong?

Thanks!

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# Displacement of spring of a block going down an incline

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