Question

A 13.0 kg stone slides down a snow-covered hill, leaving point A with a speed of 11.0 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 2.10 N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

A. What is the speed of the stone when it reaches point B?

B. How far will the stone compress the spring?

C. Will the stone move again after it has been stopped by the spring? (yes or no)

Answer 1

A)

Using law of conservation of energy  

V = sqrt(11^2 + 2 x 9.8 x 20) = 22.65 m/s

B)

Using law of conservation of energy  

0.5kx^2 = 0.5 mv^2 - umg(100 + x)

0.1l5 x 2.1 x x^2 = 0.5 x 13 x 22.65^2 - 0.2 x 13 x 9.8 x (100 + x)

1.05 x^2 = 3334.65 - 25.48(100 + x)

1.05x^2 + 25.48x - 786.65 = 0

X = 17.81 m

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