If a block is released from rest at the top of an incline and from the air from the same height, doesn’t the calculation end up the same? That is, the block starts out with its own U=mgh. The energy turns into all KE by the end so mgh=1/2(m)(v)^2, so v= sqrt(2gh). I don’t know how to do it with components (I can do it with vertical components for both but not horizontal...)
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$\begingroup$ You have already answered your own question. $\endgroup$– sammy gerbilCommented Apr 1, 2020 at 0:55
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$\begingroup$ @sammygerbil thanks it just does not make intuitive sense $\endgroup$– Alexander YeCommented Apr 1, 2020 at 0:55
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$\begingroup$ Why not? If the decrease in PE is the same while the increase in KE is different, where has the extra PE gone to? ... The block which falls reaches the ground first. It travels a shorter distance. Its acceleration is greater, but lasts for a shorter time. Perhaps this explains the confusion. $\endgroup$– sammy gerbilCommented Apr 1, 2020 at 1:07
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1 Answer
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The object will have the same speed in both cases (it doesn't matter if there is a incline or not as long as there are no dissipative forces like friction). The reason for that is the conservation of energy as you correctly pointed out. In the case of an incline, all the potential energy $U=mgh$ will be converted to kinetic energy (same as in the case without an incline) because no work can be done by the normal force.
However, you asked whether velocities are the same in both cases. The answer is no! The reason is because speed is the scalar quantity (and is the same in both cases), but velocity is the vector quantity. It is obvious that in the case without an incline the velocity will be in the vertical direction, while in the case of an incline it will have horizontal component $v \cos \alpha$ and vertical component $v \sin \alpha$, where $v=\sqrt{2gh}$ (as you concluded yourself) and $\alpha$ is the incline angle.
So to summarize my answer: speeds will be the same but velocities will not.
