Kietic friction will apply. There will be a backwards friction force opposing motion that is equal to
M g cos 5 * ìK = 1.95 m/s^2 * M
Motion will stop when the initial kinetic energy equals the gain in potential energy PLUS the work done against friction. Let X be the distance travelled. X sin 5 is the elevation gain.
(1/2) M V^2 = Mg cos5 ìK + MgX sin 5
Cancel out the M's and solve for X
V^2 = 2g [cos5 ìK + sin 5 X]
An object starts sliding up a plane tilted at an angle of alpha = 5.0◦ to
the horizontal. The object’s initial speed is v = 2.0m/s. How far from
the start point does the object come to a stop if μK = 0.20?
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