5.25-g block starts from rest at the top of a frictionless hillthat is inclined

Question

5.25-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 the block glide along the horizontal portion ofthe surface before coming to rest?
(b) If there were no friction on the horizontal surface atthe 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

Mass m = 5.25 g =5.25 * 10 ^ -3 kg

Angle   = 21 degrees

Length of the hill L = 125 m

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