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

Question

A 11.0 stone slides down a snow-covered hill (the figure (Figure 1) ), leaving point with a speed of 10.0 . 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 and then runs into a very long, light spring with force constant 2.50 . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

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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