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What is the direction of acceleration for the block moving on the xy plane?

  1. Mar 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A small block with mass 0.0400 is moving in the xy-plane. The net force on the block is described by the potential- energy function (5.80 )-(3.50 ).


    2. Relevant equations
    inverse tangent of Fy/Fx = angle of acceleration



    3. The attempt at a solution
    The question initially asked for the magnitude of the acceleration when it was at point x=0.25m y=0.63m which I derived from the equation of potential energy then finding the magnitude of the force, and finally dividing by the mass to give me the acceleration of 1.26.92 m/s^2

    The final question is asking for the direction of the acceleration of the block when it is at the same points: x=0.25m y=0.63m.

    I approached this by plugging in the magnitude of the force in the y direction (-4.167) and the magnitude of the force in the x direction (2.9) into the inverse tangent.
    So: θ=inversetangent (-4.1674/2.9) to calculate an angle of -55.167.

    It's asking for: What is the direction of the acceleration of the block when it is at the point 0.25 , 0.63 ? in degrees counterclockwise from the positive x axis. Naturally I subtracted the value for θ from 360° to get a value of 304.833°

    Unfortunately, the angle is incorrect somehow. Suggestions? Thanks!
     
  2. jcsd
  3. Mar 15, 2012 #2
    How did you do the first part exactly? What is the magnitude of the acceleration of the block when it is at the points?
     
  4. Mar 15, 2012 #3
    Considering the equation for the potential energy is U(x,y)=(5.8J/m2)x2-(3.50J/m^3)y^3 that makes the derivative of that equation 11.6(x) -10.5y2

    Then I plugged in the coordinates the problem provided (x=0.25m, y=0.63m) and calculated Fx=2.9N and Fy=-4.16745N

    Then to get the magnitude of the F I used Pythagoras. |F|=√2.92+ -4.167452 which gave me a magnitude of 5.077 N

    Because F=ma I divided 5.077N by the mass of 0.0400 kg and got the acceleration of 126.92 which was verified as correct.
     
  5. Mar 15, 2012 #4
    My equation is U=5.65x^2 - 3.6y^3, mass = .04kg, x = .3m, y = .52m:

    i took the partial derivative of U with respect to x and with respect to y.

    Fx = -dU/dx = -11.3x
    Fy = -dU/dy = -10.8y^2

    using the same method that you did F = sqrt(Fx^2 + Fy^2) i got F = 4.474

    F=ma so 4.474/.04 == 111.86 m/s^2


    for the second part I thought it would be tan^-1(Fy/Fx) === 40.74 degrees

    but this did not work.
     
  6. Mar 15, 2012 #5
    I dont know if it will help you but I found a problem in my physics book that has the equation U(x,y) = 5.8x^2 - 3.6y^3, mass is .04 kg, x = .3m, y = .6m.

    Correct Answers:

    F = 130 m/s^2

    132 degrees counterclockwise from the +x axis
     
  7. Mar 15, 2012 #6
    I solved the problem I posted and found that if you take 180 - (tan^-1(Fy/Fx)) you will get the correct answer.
     
  8. Mar 15, 2012 #7
    That answer also worked in mastering physics.
     
  9. Mar 15, 2012 #8
    when I take inverse tangent of (-4.16745/2.9) (y/x) I get -55.167, since its asking for the angle counterclockwise from the positive x axis I added 360, which gave me an angle of 304 degrees, masteringphysics says this is incorrect.

    I've also tried entering 180-inversetangent(-4.16745/2.9) and get an answer of 235 degrees, which mastering physics also says is wrong.

    Just for kicks I've also tried entering the angle of -55 which would be clockwise from the positive x axis, this is also wrong.

    I'm at a loss.
     
  10. Mar 15, 2012 #9
    Update, just for kicks I tried adding 180 to the angle of -55 degrees. I calculated 124.83 (mostly drawing on inspiration from what you said was in your physics book and trying to scale my answer accordingly) and HA! Masteringphysics graded this as correct.

    My only problem now is understanding how this makes since.
     
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