A 11.0 kg stone slides down a snow-covered hill (the figure (Figure 1) ), leavin

Question

A 11.0 kg stone slides down a snow-covered hill (the figure (Figure 1) ), leaving point with a speed of 12.0 m/s. There is no friction on the hill between points and , but there is friction on the level ground at the bottom of the hill, between 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 2.50 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

Since you didn't state what the height of A above B is, I can only assume that the velocity at B is also 10 m/s parallel to the level ground... Taking gravity to be 10(m/s)/s, Normal force = 11 * 10 = 110N Since the stone is moving, friction force is 0.20 * 110N = 22N. a = F / m = 22N / 11.0kg = 2(m/s)/s (v(f))^2 = (v(i))^2 + 2ad = 121 + 2(-2)(100) = 121 - 400 = -279. In conclusion, we really need the height of A from the level ground

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