A box on a frictionless lamp?

[m25]

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Say we have a frictionless ramp with length 3 and vertical height h with a box at the top of the slide, and another one of the exact same system except the ramp length is 5, and have both of the boxes slide down the ramp. The final velocity of the two boxes will be the same since we are simply converting all the gravitational potential energy(mgh) into kinetic energy, since the height is h for both of them.
Now, my question is, does it mean that they would also take the same amount of time to slide down the ramp?

 

[justadream]

@m25

No, I think the longer the ramp, the longer the time necessary to reach the bottom.

If you keep height constant but lengthen the ramp, the force the box feels due to gravity (mgsin(theta) ) will decrease.

F = ma
if Force decreases, acceleration must decrease.

Thus, time must increase.


Alternatively, just make an extreme example. Think about if the ramp reached .5m in height but was 99999999999999999999999999999m long. Now compare that to a ramp that also reached .5m in height but was .00000000001m long.

In the second scenario, you'd reach the bottom in a second or less. In the first scenario, even if you could move the box at the speed of light, you wouldn't see the box move to the bottom in a second or so.

 

[Cawolf]

Right, both will have a velocity at the bottom of v = sqrt(2gh).

As @justadream said - the longer the ramp (with constant height), the smaller the angle, the weaker the horizontal component of gravity.

 

[techfan]

Edit: bad explanation