What is the deceleration on an incline?

AI Thread Summary
To calculate the deceleration of an object moving up an incline, one must consider the forces acting on it, particularly the gravitational force component along the incline, which is -mg sin(theta). In a specific example, an object entering a 25-degree incline at 0.25 m/s was analyzed using energy equations, leading to a calculated vertical height of approximately 3.2 millimeters. The length along the incline was determined to be about 7.55 millimeters using the correct equations. It was noted that using consistent variable names is crucial to avoid confusion in calculations. Overall, the discussion emphasizes the importance of understanding forces and proper notation in physics problems involving inclines.
KingNothing
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Hey all, I have a question. How can I calculate the deceleration of an object that begins to go up an incline of angle X? See there are a lot of problems like this. One of mine for example involves an object entering an incline of 25 degrees at 0.25 m/s, and I need to calculate how far it goes up the incline.
 
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KingNothing,

Draw a diagram of the incline and the object. What direction would a force have to be in order to decelerate the object? Draw vectors for the forces acting on the object. Which one(s) of these could contribute to decelerating the object?
 
I have...I get gravity coming from the top, force normal going perpendicular to the hypotenuse side...and that's all (there is no friction in this problem).
 
If there is no friction, the force normal to the hypotenuse is irrelevant. The force is the component of force ALONG the hypotenuse. That would be -mg sin(theta) where theta is the angle the slope makes with the horizontal.
 
I think I got this one wrong...instead of force equations, I used energy equations...

If the speed is .25 m/s when it enters the incline, then I set potential energy equal to kinetic energy...
.5m(.25^2)=mgh
.5*.25^2=9.8h

or about 3.2 millimeters, which would be the vertical component of the incline, so...
l=\frac{.5*.25^2}{9.8*sin25} or about 7.55 millimeters.

I did use that method, at one point...if you do that, and use a mass of 1 kg for simplicity, you get d=.5(9.8*sin25)(.25/(9.8*sin25))^2 or the same answer.
 
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Hi,

The equation for the length up the incline is correct (I didn't check the numbers, but I assume they're ok too.). The only thing I would suggest is that you not use a single variable for more than one thing. You use h to mean height off of the ground in your first equation and then you use h to mean length along the incline in your last equation. I was able to follow what you did, but you're liable to confuse yourself and others if you do things like that. You've got 26 letters, capital and lower case, all the greek alphabet, and all the subscripts you want to make up new variable names; it's worth it to do so.
 
jamesrc said:
Hi,

The equation for the length up the incline is correct (I didn't check the numbers, but I assume they're ok too.). The only thing I would suggest is that you not use a single variable for more than one thing. You use h to mean height off of the ground in your first equation and then you use h to mean length along the incline in your last equation. I was able to follow what you did, but you're liable to confuse yourself and others if you do things like that. You've got 26 letters, capital and lower case, all the greek alphabet, and all the subscripts you want to make up new variable names; it's worth it to do so.

Whoops, that was a mistake with the last h. I usually do use pretty decent/legitbble notation.
 
OK. Didn't mean to get on a soapbox there; mistakes happen.
 
It's all good:). Thanks James.
 
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