A block of mass 0.540 kg is pushedagainst a horizontal spring of negligible mass

Question

A block of mass 0.540 kg is pushedagainst a horizontal spring of negligible mass until the spring iscompressed a distance x. The force constant of the springis 450 N/m. When it is released, the block travels along africtionless, horizontal surface to point B, the bottom ofa vertical circular track of radius R = 1.00 m, andcontinues to move up the track. The speed of the block at thebottom of the track is vB = 13.1 m/s, and the block experiences an averagefrictional force of 7.00 N while sliding up the track.

Explanation / Answer

(a)What is x?
   nc = KE +Ug + Us
      nc =(1/2)*m*vf2 -(1/2)*m*vi2 + mghf -mghi + (1/2)*k*xf2 -(1/2)*k*xi2
         nc= (1/2)*m*vf2 -(1/2)*k*xf2
         0 =(1/2)(0.540 kg)(13.1 m/s)2 -(1/2)(450)(xf)2
         x =0.206 m

(b)What speed do you predict?    nc = KE +Ug + Us
      nc =[(1/2)*m*vf2 -(1/2)*m*vi2] + [mghf -mghi] + [(1/2)*k*xf2 -(1/2)*k*xi2]
         nc= (1/2)*m*vf2 + mghf +(1/2)*k*xf2
         7.00* =(1/2)(0.540 kg)(vf2) + (0.540 kg)(9.81m/s2)(2 m) + (1/2)(450)(0.206 m)2
         vf= 12.5 m/s

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