Incline Planes: Final velocity and Time

[September24]

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According to a specific question, when a ball is dropped down inclined plane vs simply dropped from a height H, the final velocity is the same but time is longer when on incline plane. I thought that for incline plane, gravity would be a fraction of what is usually is so final velocity AND time would be affected (v=at)

 

[Heday]

According to a specific question, when a ball is dropped down inclined plane vs simply dropped from a height H, the final velocity is the same but time is longer when on incline plane. I thought that for incline plane, gravity would be a fraction of what is usually is so final velocity AND time would be affected (v=at)

Let's think of the ball in terms of energy.

PE = mgh

If a ball falls from height H OR is rolled down an inclined plane that starts at height H, both balls will have the same PE; we can agree that when it reaches the bottom, it will have zero PE because all of the initial PE would be converted to KE, right? So we're saying if mgH is the PE for both balls, and it is zero for both balls when it reaches the ground, then we can says that KE(dropped)=KE(rolled). KE=0.5mv^2, if the masses of the balls are equal then velocity would therefore need to be equal.

You are right to say that gravity would only be a fraction of what is normally is, on an inclined plane. Therefore, if we're using h=0.5at^2, "h" is equal for both balls, however "a" would be smaller for the ball rolling down the inclined plane, requiring "t" to be bigger.

 

[Temperature101]

According to a specific question, when a ball is dropped down inclined plane vs simply dropped from a height H, the final velocity is the same but time is longer when on incline plane. I thought that for incline plane, gravity would be a fraction of what is usually is so final velocity AND time would be affected (v=at)

Don't the specific question but when the ball dropped from a height, it will take less time than if it was dropped down an inclined plane... Let say, the time it that from an height H is 2s and acceleration is 10 m/s^2... Therefore, v = 20 m/s.... If it is on an incline plane with an angle of 45 degree for instance, acceleration might be 5 m/s^2 and time might be 4 seconds... V will still be 20 m/s^2... As the angle get smaller, the ball takes more time...

 

[September24]

Okay so....using the V=AT equation

While acceleration does get smaller, time increases so they cancel each other out?

In terms of conservation of energy, it makes senses since both balls have the same PE in the beginning and same PE at the end.

 

[Heday]

Okay so....using the V=AT equation

While acceleration does get smaller, time increases so they cancel each other out?

In terms of conservation of energy, it makes senses since both balls have the same PE in the beginning and same PE at the end.

Yes. It's all interrelated.