An inclined plane of angle = 20.0° has a spring of force constant k = 535 N/m fa

Question

An inclined plane of angle = 20.0° has a spring of force constant k = 535 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 = 2.45 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?

please show method, do not give me the same answer I got

Your answer: 0.11107 is more then 10% incorrect

Explanation / Answer

Apply the work energy thorem

Ki+ Wg + Ws = Kf

1/2 mvi^2 + mg sin theta( d+ x) + (1/2 k xi^2 - 1/2 k xf^2) = 1/2 mvf^2

1/2 mvf^2+ mg sin thets ( d+ x) + (0-1/2 kx^2) = 0

divide with m

1/2 vf^2+ g sin thets ( d+ x) + (0-1/2 k/mx^2) = 0

k/2m x^2 - ( g sin tehta) x - ( v^2/2 + ( g sin tetha) d) = 0

solving

x = g sin tehta + sqrt ( g sin tehta)^2 + ( 2k/m) ( v^2/2 + g d sin theta)/k/m

= 9.8 sin20 + sqrt ( 9.8 sin 20)^2 + ( 2(535)/2.45( (0.750)^2/2 + 9.8(0.318) sin 20/535/2.45

=0.127 m

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