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Find the magnitude of the block's acceleration

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Refer to Fig. 5-38 (see below). Let the mass of the block be 2.3 kg and the angle x be 20°. Find (a) the tension in the cord and (b) the normal force acting on the block. (c) If the cord is cut, find the magnitude of the block's acceleration.


    2. Relevant equations
    F=ma (Newton's Second Law)


    3. The attempt at a solution

    I got a and b by figuring out the equations T=F(g)sinx and F=F(g)cosx where F(g) is the force due to gravity. Part C is where I have the problem. I thought that because the rope was cut, the tension there would be 0 and thus the acceleration would be the tenstion
     

    Attached Files:

  2. jcsd
  3. Feb 5, 2008 #2

    Doc Al

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    The tension is zero--the string no longer pulls on the block. Once the string is cut, what's the net force on the block? (What forces act on the block?)
     
  4. Feb 5, 2008 #3
    Right, thats what i figured. So wouldnt the net force be in the direction of opposite of the tension (ie in the picture, it would be F(g)sinx?)
     
  5. Feb 5, 2008 #4

    Doc Al

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    Exactly! The only unbalanced force on the block is the component of gravity down the incline, which is [itex]F_g \sin\theta[/itex].
     
  6. Feb 5, 2008 #5
    Right, but since its an online program...i put in 7.71 and -7.71 and neither is right. The force due to gravity is 22.54 N and multiply that by sin20 = 7.71, but its apparently not right?
     
  7. Feb 5, 2008 #6

    Doc Al

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    They want the acceleration, not the force. You're not done yet.
     
  8. Feb 5, 2008 #7
    So, since thats the force...F=ma? would you divide that by the mass then?
     
  9. Feb 5, 2008 #8

    Doc Al

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    Staff: Mentor

    Right.
     
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