The block shown in the figure has mass =7.0 kg and lies on a fixed smooth frictionless plane tilted at an angle theta=22.0 degrees to the horizontal.
(a) Determine the acceleration of the block as it slides down the plane.
(b) If the block starts from rest 12.0 m up the plane from its base, what will be the block’s speed when it reaches the bottom of the incline?
Answer:
For this block the force acting along the direction of the slope is mgsin22o acting in the downward direction along the slope. Since this is the only force acting along this direction,
mgsin22 = ma
=> a = gsin22 = 3.671 m/s2
b] Initial velocity = 0 m/s
Distance covered = S = 12 m
use, v2 = u2 + 2aS
so, v = [0 + 2(3.671)(12)]1/2 = 9.386 m/s
this is the speed of the block at the bottom.
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