Velocity down an incline

  • Thread starter keithcuda
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  • #1
keithcuda
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Homework Statement


A 75KG object, goes down a slide that is 10m tall. What is the speed of the object at the halfway mark and the speed at the bottom of the slide. The object starts at rest and there is no friction.

Homework Equations


v2f - v2i = 2as


3. The Attempt at a Solution [/B]
attached pic of where I am at on the problem
incline:velocity.jpg
 

Answers and Replies

  • #2
ehild
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You have the equation: v2f - v2i = 2as. Note that s is the displacement along the slope now. How long is it in terms of the angle if the height of the slope is 10 m?
 
  • #3
keithcuda
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You have the equation: v2f - v2i = 2as. Note that s is the displacement along the slope now. How long is it in terms of the angle if the height of the slope is 10 m?
sinθ=(10/h)
h equaling the hypotenuse = the slope
 
  • #4
ehild
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No, h=10 m, and s is the hypotenuse. Look at the picture.
 
  • #5
keithcuda
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No, h=10 m, and s is the hypotenuse. Look at the picture.
ok so,

sin(θ)=(10/s)
 
  • #6
ehild
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Isolate s and use it in the equation v2f - v2i = 2as.
 
  • #7
keithcuda
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so to isolate s, do you mean something like this?

s=10sin(θ)
 
  • #8
ehild
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so to isolate s, do you mean something like this?

s=10sin(θ)
Something like that, but not quite. Is it more clear if I say solve the equation sin(θ)=(10/s) for s?
 
  • #9
keithcuda
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Something like that, but not quite. Is it more clear if I say solve the equation sin(θ)=(10/s) for s?
more like this?
s=10/sin(θ)

I'm sorry, this stuff has me so confused... These online classes aren't really meeting my expectations. :(
 
Last edited:
  • #10
ehild
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Yes.
Determine the acceleration a now.
 
  • #11
keithcuda
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Yes.
Determine the acceleration a now.
well, acceleration = gsin(θ)
so, (9.81)sin(θ)=a
 
  • #12
ehild
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Use a and s in the formula v2f - v2i = 2as.
 
  • #13
keithcuda
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well v2i = 0 since the object starts at rest
v2f = 2((9.81)sin(θ)) (10/sin(θ))
 
  • #14
ehild
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Good! Evaluate.
 
  • #15
keithcuda
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I'm not sure if this is right, but it's a shot in the dark...

v2f =2 ((9.81)sin(θ) (10/sin(θ))
umm??? sin(θ) cancels out on each side leaving, 2(9.81)(10) = 196.2
v2f = 196.2
vf = √196.2 = 14m/s

is this even close?
 
  • #16
ehild
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It is correct. That is the speed at the bottom of the slope when the whole length is covered. What is the speed, when the object is halfway down the slope?
 
  • #17
keithcuda
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It is correct. That is the speed at the bottom of the slope when the whole length is covered. What is the speed, when the object is halfway down the slope?
lol i would say half of that? Thats just me guessing...
 
  • #18
ehild
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lol i would say half of that? Thats just me guessing...
NO!!!!!!!!!!!!! what is the new distance s(half) if it is half the previous one?
 
  • #19
keithcuda
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NO!!!!!!!!!!!!! what is the new distance s(half) if it is half the previous one?
haha I knew it wouldn't be that easy ;)

v2f = 2 * 9.81(10 * .5)
vf = √98.1
vf = 9.9m/s
 
  • #20
ehild
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You did it!
 
  • #21
keithcuda
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I feel the need to celebrate!!!
Thank you very much!

When I started this class, I really thought it would be one of my favorite subjects. I like it, but it really has me scratching my head.

I greatly appreciate it!
Merry Christmas!!

Cheers!
 
  • #22
ehild
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Merry Christmas to you, too. :)
 

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