Why does increasing the angle of an inclined plane not affect the velocity at the bottom?

[deleted647690]

See above

Is it just that it doesn't change the amount of potential energy the object has at the top?

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[theonlytycrane]

Assuming the box stays at the same height regardless of angle, the initial potential energy mgh will stay the same. With conservation of energy, the final kinetic energy at the bottom will in turn stay the same (as you described).

 

[aldol16]

Is it just that it doesn't change the amount of potential energy the object has at the top?

On a frictionless inclined plane, the angle of incline doesn't affect the final velocity because of energy conservation. But it's important to note that it's for a frictionless incline (which is most likely what you will encounter on the MCAT, if at all).

 

[victorias]

If changing the angle of the incline doesn't affect the height of the plane, then the velocity at the bottom will not change because there is still the same potential energy at the top which is totally converted to kinetic energy at the bottom based on the Law of Conservation of Energy and this only applies to a frictionless plane. But changing the angle will change the velocity in transit as the object is sliding down the plane which is mg sin (theta). So if you increase the angle of the plane, sin (theta) increases and thus velocity at which the object slides down the plane also increases.

@aldol16 Can you go over an example with friction?

I am thinking that if there is friction, then we can set up the equation as PE = KE + Friction since some of the potential energy will be lost to friction and will not be fully converted to kinetic energy so the maximum possible velocity will not be reach when the object reaches the bottom. Is my thinking correct?

 

[aldol16]

@aldol16 Can you go over an example with friction?

I am thinking that if there is friction, then we can set up the equation as PE = KE + Friction since some of the potential energy will be lost to friction and will not be fully converted to kinetic energy so the maximum possible velocity will not be reach when the object reaches the bottom. Is my thinking correct?

Conceptually, that sounds right but I'm not exactly sure. I haven't done classical mechanics in awhile. In terms of the work-energy theorem, you'd just have W = conservative W + nonconservative W = KE. So yeah, you should be able to figure out what work friction does on the block as it slides down and subtract that from the energy at the bottom.

Changing the slope will change the velocity at the bottom because you change the normal force on the block. Changing the normal force necessarily changes the frictional force exerted on the object. So why don't you do an example where you incorporate friction with an angle of, say, 30 degrees and a coefficient of kinetic friction of 0.4?

 

[wizzed101]

If you want to think about this intuitively, you can imagine that increasing the plane will indeed increase the acceleration however, the time it takes for the object to reach the bottom will also diminish. So on one hand, you will have big acceleration (large angle) but little time to reach terminal velocity, and on the other, there is a much smaller acceleration but plentiful time.

Frictionless or not, it does make sense.

 

[Mcoles]

When seeing this problem from the perspective of conservative forces i.e. frictionless plane, if the height that the object drops down the ramp from doesn't change then its initial potential energy doesn't change no matter what the angle. Conservative forces imply that the path it takes doesn't matter. Thus, Initial Potential Energy (mgh) = Final Kinetic Energy (1/2mv^2). Nothing related to the angle of the incline in that equation.

 

[deleted647690]

So the larger incline plane will increase at a greater rate (a greater acceleration), but will reach the same final velocity as the lower incline plane, because the lower incline plane has a lower acceleration but longer time accelerating to reach that final velocity

 

[desperadoflakes]

Assuming the plane is frictionless, the only force acting on the box is gravity. Because gravity works in the y-direction, only the y-velocity will change. So to figure out how fast the ball is moving, we need to ask two questions. 1) How much gravitational force is acting on the object? 2) How long is the gravitational force acting on the object for?

We can collate these two questions into one question: How much work is being done on the object by gravity?
The amount of work being done on the object is the same as the gravitational potential energy (U) the box has at the top of the ramp.
U = mgh , where h is the vertical distance from the top to the bottom of the ramp

So as long as h (the vertical distance from top to bottom) remains the same, the work being done by gravity will be the same, and the total kinetic energy at the bottom of the ramp will be the same, and the velocity will be the same. When the ramp is more steep, there is a larger gravitational force acting on the object (mg * sin theta). So the answer to question 1 is more gravitational force. However, because it will pick up speed more quickly, it will reach the bottom of the ramp more quickly, which means the overall time of travel is shorter. So the answer to question 2 is less time. So it cancels out.

The key thing to understand here is that "h" stays the same. Because "h" = Lsintheta, this also means that if h stays the same while theta changes, that means that the length of the ramp must also change. If theta goes up, L must go down for h to be the same.