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

g = 9.80 m/s^{2}.

How far up the incline (measured parallel to the surface of the incline) does the block travel before it starts to slide back down?

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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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