A 15.0 kg stone slides down a snow-covered hill, leaving point A with a speed of

Question

A 15.0 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. How far will the stone compress the spring?

Explanation / Answer

The speed when it reaches the B depends on the vertical height alone

u² = 12² +19.6*20 (using the equation v² = u² + 2as)

u = 23.15 m/s

The acceleration due to friction is -g.

The velocity after travelling 100 m is given by

v² = u² - 2as = 536 - (2*0.2*9.8*100) = 144. (Using kinetic friction since it is in motion)

If the spring is compressed to a distance s, (Using energy principle)

v² = 2gs + (k/m) s²

144 = (2*0.2*9.8*s) + (2.2 /15) s²

s = 36.69 m

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