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milliex51

  • 4 years ago

How do I calculate the acceleration of the cart shown, given the following assumption: A frictional resistance of magnitude 2.0 N is acting on the cart.

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  1. milliex51
    • 4 years ago
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    |dw:1324423827086:dw|

  2. henkjan
    • 4 years ago
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    F = m*a F = 2 N m = 6 kg??? (not really clear what the cart exactly is from your drawing) F = m*a a = F/m a = 2/6 = 1/3 m/s2 (a deceleration, maybe you want to add a minus sign in the force of the resistance)

  3. milliex51
    • 4 years ago
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    The answer is 2.9 m/s ^2 [fwd] in the back of my book.. but

  4. henkjan
    • 4 years ago
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    than i misinterpretted your picture, let me guess again :P

  5. milliex51
    • 4 years ago
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    and I found the formula for no friction only: T = m1a a = m2g/(m1 + m2)

  6. hyunaelee
    • 4 years ago
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    F=MA. Which means summation of forces is equal to mass times acceleration. There are two different blocks so there are two different free body diagrams. Block of 4 KG. There's a force normal and a force gravity and a force friction and a force tension. Block of 2 KG. There's a force tension and a force gravity only. So for the 2nd block Ft-Fg= MA. --> Ft-2.0(9.8)=2.0(a) ---> 4.0a+2.0-2.0(9.8)=2.0a solve for a 1st block. Ft-Ff=MA. ---> Ft-2.0 = 4.0(a) Ft=4.0a+2.0

  7. TuringTest
    • 4 years ago
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    m=4kg from her pic

  8. milliex51
    • 4 years ago
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    ohhh.. I think I'm getting it. o.o

  9. milliex51
    • 4 years ago
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    May I ask why ft - Fg = ma?

  10. milliex51
    • 4 years ago
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    sometimes I don't get these formulas >_<"

  11. hyunaelee
    • 4 years ago
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    Look at the block. Look at what's "touching it" And then there are also 4-non contact forces that can act upon the block. Force gravity, electric force, magnetism force, and I forget the last one

  12. milliex51
    • 4 years ago
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    yes, tension force? normal force? applied force?

  13. hyunaelee
    • 4 years ago
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    So for the 2.0 KG block only a rope is touching the block. Which is Force Tension (Ft) and a non-contact force force gravity (Fg)

  14. cwrw238
    • 4 years ago
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    equtoion of motion for 2.5kg is 2.5g - T = 2.5a ......................................4kg is T - 2 = 4.5a solving this sytem will give you the reqd acceleration

  15. TuringTest
    • 4 years ago
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    because total force=ma and the total force is force of the object pulling on one end of the string minus the friction force (because it is acting in the opposite direction).

  16. milliex51
    • 4 years ago
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    yup, I see it! :)

  17. milliex51
    • 4 years ago
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    ohhhhhhhh.

  18. milliex51
    • 4 years ago
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    thank you!!

  19. hyunaelee
    • 4 years ago
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    No problem. I hope you understand the problem now ^_^. Remember to draw free body diagrams of the entire system and remember contact forces and non-contact forces

  20. milliex51
    • 4 years ago
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    yeah I will! Wait, what do you mean by contact and non-contact forces? Like tension force, gravity force, normal force, applied force... etc?

  21. TuringTest
    • 4 years ago
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    gravity force is non-contact because it touches nothing in order to impart a force on it.

  22. hyunaelee
    • 4 years ago
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    Contact is forces that are literally touching the object you are drawing the free body diagram for. Non contact forces are "invisible" forces. There are only 4 forces that are "invisible" but I think the only one you'll ever face in basic physics is "Force Gravity". There are two more for more advanced physics "Electrical and Magnetism". Contact forces can be : Force friction (From a surface) or force tension (from a rope) force spring (from a spring) and much more

  23. milliex51
    • 4 years ago
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    ohhhh, i see, i see. THANK YOU S O MUCH!! :D

  24. cwrw238
    • 4 years ago
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    you sure its 2.9 m/s/s for acceleration

  25. hyunaelee
    • 4 years ago
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    No problems. Good luck ^_^ and hope you do well.

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