A block of mass m = 39 kg is slid over a horizontal frictionlesssurface at some

Question

A block of mass m = 39 kg is slid over a horizontal frictionlesssurface at some constant velocity. It then encounters a inclinedplane which has a coefficient of kinetic friction of 0.39. Theinclination angle for the incline is x = 34.6 degrees and theheight of the incline is h = 23.8 meters above its base. If themagnitude of the velocity at the top of incline is 34 m/s for theblock, what was the magnitude of its velocity at the bottom of theincline in m/s?

The image can be found at the link below.

http://i234.photobucket.com/albums/ee269/yankeekd25/ramplauch.gif

Explanation / Answer

let the speed at bottom be vi, vf = 34m/s, let acceleration be a, we have: -mgcos(34.6) - mgsin(34.6) = ma a = -g(cos(34.6)*0.39+sin(34.6)) = -8.70m/s2 s = h/sin(34.6) = 41.9 m since vf2 - vi2 =2as you can get vi = [vf2-2as] = 43.42 m/s s = h/sin(34.6) = 41.9 m since vf2 - vi2 =2as you can get vi = [vf2-2as] = 43.42 m/s

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