A block of mass 0.570 kg is pushed againsta horizontal spring of negligible mass

Question

A block of mass 0.570 kg is pushed againsta 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 = 13.6 m/s, and the block experiences an averagefrictional force of 7.00 N while sliding up the track.

Explanation / Answer

Applying the law of conservation of energy we have as, KEat B=(KE+PE+fr)at the top 1/2*mvB2=1/2*mv2+7d+mgh 1/2*0.57*13.62=1/2*0.57*v2+7*2*1+0.57*9.8*2*1 v2=27.5416 v=5.2480091m/s Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework."

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