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Frictionless Inclined Plane, solve for acceleration

  • Thread starter lexikobie
  • Start date
  • #1
7
1

Homework Statement


An object is on a frictionless inclined plane. The plane is at an angle of 30 degrees with the horizontal. What is the object's acceleration?

a) 0.50g b) 0.56g c)0.68g d)1.0g e)0.87g

Homework Equations


F=ma a = g.sin(theta)

The Attempt at a Solution


I set it up where a = mg.sin(theta) / m where I then get a = g.sin(theta)
where it's: 9.8 (sin(30)) and my acceleration would be 4.9 m/s^2. However, that is
not an answer choice with my options given, where I am now completely lost.
I sincerely would appreciate any help!
 

Answers and Replies

  • #2
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For the answers, they are a number multiplied by g.
 
  • #3
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For the answers, they are a number multiplied by g.
So when I set up the equation to find the acceleration which I thought was 4.9, was that the wrong thing to do? What am I missing because I know this is an easy problem, but I'm having trouble with this...
 
  • #4
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You've said that your answer isn't in the options given, you should take another look at the options and consider what I posted above.
 
  • #5
7
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You've said that your answer isn't in the options given, you should take another look at the options and consider what I posted above.
Ohhhhh..... Okay wow, then if that's the case I believe it is answer choice A. Because the sin of 30 is 1/2. And for my acceleration, it's gravity x sin(30). That's what you meant right? I did solve it correctly after all I suppose^^
 
  • #6
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That's exactly it. :) With answers like those it's a good idea to either convert the answers so that they're not in terms of g (So a is 0.5g = 0.5 * 9.8 = 4.9) or to get your answer in terms of g (sin(30)g = 0.5g).
 
  • #7
7
1
That's exactly it. :) With answers like those it's a good idea to either convert the answers so that they're not in terms of g (So a is 0.5g = 0.5 * 9.8 = 4.9) or to get your answer in terms of g (sin(30)g = 0.5g).
Thank you so very, very much Mike! Have a Blessed one!
 
  • #8
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Glad to help!
 

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