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An inclined plane of angle θ has a spring of force constant k fastened securely at the bottom so that the spring is parallel to the surface. A block of mass m is placed on the plane at a distance d from the spring. From this position, the block is projected downward toward the spring with speed v as shown in the figure below. By what distance is the spring compressed when the block momentarily comes to rest? (Use any variable or symbol stated above along with the following as necessary: g, the acceleration due to gravity.)

Respuesta :

zoexoe
While assuming a frictionless inclined plane, we use the work-energy theorem W = ΔE:
       Ki + Wg + Ws = Kf, showing that the mechanical energy of the system is conserved        
 
       (1/2)mv2 + mgsinθ(d+x) + [0 - (1/2)kx2] = 0 where the vertical height from the final position is sinθ(d+x).

To simplify, we divide the equation by m:
       (1/2)v2 + gsinθ(d+x) - (k/2m)x2 = 0
       (k/2m)x2 - (gsinθ)x - (gsinθ)d - (v2/2) = 0

Arranging the quadratic equation in the form ax2 + bx + c = 0,
       (k/2m)x2 - (gsinθ)x - [ (v2/2) + (gsinθ)d ] = 0

we can solve for x using the quadratic formula:
       x =  {gsinθ ± sqrt[ (gsinθ)2 + 4(k/2m)(v2/2 + (gsinθ)d) ]} / 2(k/2m)
       x =  {gsinθ + sqrt[ (gsinθ)2 +  (2k/m)(v2/2 + gdsinθ) ]} / (k/m)

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