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A block of mass 3 kg slides down a frictionless inclined plane of length 6 m and height 4 m. If the block is released from rest at the top of the incline, what is its speed at the bottom?
(if it's frictionless the length doesn't even matter :) ) It would have the same kinetic energy down as the potential energy up. That is, [tex]mgh=\frac{mv^2}{2}[/tex] or [tex]2gh=v^2[/tex] (the mass doesn't even matter). The result is [tex]\sqrt{2gh}[/tex], so only the height matters really. It is almost 9 (it is [tex]\sqrt{80}=4\sqrt{5}[/tex]).
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