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anonymous

  • 4 years ago

A 1.5-kg cart is pulled with a force of 7.3 N at an angle of 40 degrees above the horizontal. If a kinetic friction force of 3.2 N acts against the motion, the cart's acceleration along the horizontal surface will be?

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  1. anonymous
    • 4 years ago
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    This is a simple balance of forces.\[\sum F = m*a = F - F_f\]

  2. anonymous
    • 4 years ago
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    what is the F=? and Ff=?

  3. anonymous
    • 4 years ago
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    i don't get it..

  4. anonymous
    • 4 years ago
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    Okay. From Newton's first law, The net force equals the mass times the acceleration. F is the applied force, F_f is the frictional force. Since frictional force always opposes the motion, it must be directed in the opposite direction as the applied force.

  5. anonymous
    • 4 years ago
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    Newton's second law rather.

  6. anonymous
    • 4 years ago
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    yes. Oh.. so it's Fa - Ff = m*a = F

  7. anonymous
    • 4 years ago
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    what's the angel's role?

  8. anonymous
    • 4 years ago
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    It is just\[m*a = F_a - F_f\]

  9. anonymous
    • 4 years ago
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    ohh.

  10. anonymous
    • 4 years ago
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    Ah. I misread the problem. We need to find the component of the applied force that is directed along the path of motion. \[F_p = F_a \sin(\theta)\]\[m*a = F_p - F_f\]

  11. anonymous
    • 4 years ago
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    what is Fp?

  12. anonymous
    • 4 years ago
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    I told you what F_p is. \[F_p = F_a \sin(\theta)\]It is the component of the force that acts along the path of motion.

  13. anonymous
    • 4 years ago
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    oh but like is it like applied force..

  14. anonymous
    • 4 years ago
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    Fp = 7.3 N sin(40 = 5.44 m*a = Fp - Ff 1.5*a = 5.44 - 3.2 1.5*a = 2.24 a = 1.49?

  15. anonymous
    • 4 years ago
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    Yes.

  16. anonymous
    • 4 years ago
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    Do you understand why?

  17. anonymous
    • 4 years ago
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    no, I don't. Plus it's not in the answer provided.. please explain D: I'm so bad at physics i don't even know why i took it >_<"

  18. anonymous
    • 4 years ago
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    First, begin every problem like this with a free body diagram. |dw:1327287228367:dw|Now, we need to break up the applied force into its normal components. That is, x and y based on the drawn coordinate system (a coordinate system should be included in every free body diagram)\[F_x = F_a \cos(\theta)\]\[F_y = F_a \sin(\theta)\](Whoops! I gave you the wrong expression for the applied force. I'm sorry.) We are interested in the horizontal acceleration. Or the acceleration in the x-direction. Now, let's write a balance of forces. \[m*a_x = F_a \cos(\theta) - F_f\]

  19. anonymous
    • 4 years ago
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    ohh.. I'm kind of getting it.. thank you!!

  20. anonymous
    • 4 years ago
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    ahh.. m*ax = Facos(theta)-Ff 1.5 * ax = 7.3cos(40)-(-3.2) ax = -1.67/1.5 ax = -1.113?

  21. anonymous
    • 4 years ago
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    1.5 * a_x = 7.3 cos(40) - 3.2 Watch you signs. From the above free body diagram, we can see that F_f opposes the applied force and the resulting motion. Therefore, when setting up the force balance equation, we already account for it being negative.

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