(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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Displacement of spring of a block going down an incline

**Physics Forums | Science Articles, Homework Help, Discussion**