A small block travels up a frictionless incline that is at an angle of 30.0° above the horizontal. The block has speed 4.21 m/s at the bottom of the incline. Assume
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THE NET acceleration acting on the block is = (- g Cos 60)
so, u = 4.21m/s
a = (- g Cos 60) = -9.8 x 0.5 = -4.9m/s
v = 0 m/s
applying v^2 = u^2 + 2aS
where S is the distance covered.
on calculating S= 1.80 m
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