A 11.0 kg stone slides down a snow-covered hill , leaving point A with a speed o
ID: 1344450 • Letter: A
Question
A 11.0 kg stone slides down a snow-covered hill , leaving point A with a speed of 12.0 m/s . There is no friction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100 m and then runs into a very long, light spring with force constant 2.20 N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.
What is the speed of the stone when it reaches point B?
Explanation / Answer
h = height of hill given in the diagram
using conservation of energy between point A and B
PE at A + KE at A = KE at B
mgh + (0.5) m Va2 = (0.5) m Vb2
Vb2 = 122 + 2 x 9.8 x h
Vb = sqrt (144 + 19.6 h)
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