A5.25-g block starts from rest at the top of a frictionless hillthat is inclined

Question

A5.25-g block starts from rest at the top of a frictionless hillthat is inclined 21.0º with respect to the horizontal. The hill is125 m long. At the bottom of the hill, the surfaceis level and the coefficient of kinetic friction between thesurface and the block is now 0.00655. (a) How far does theblock glide along the horizontal portion of the surface beforecoming to rest? (b) If there were no friction on thehorizontal surface at the bottom of the hill, but instead a spring(k = 1250 N/m) was located 15.0 m along the horizontalsurface from the bottom of the hill, how far would the spring becompressed when the block strikes it before the block comes to astop?

Explanation / Answer

a) loss in potential energy =work done by friction force; mgh=mgcos*125+mg*x;        x=(125cos21-125cos21*0.006955)/0.00655 x=0.890m

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