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Finding force on a lab cart

  1. Jul 16, 2009 #1
    1. When a student exerts a force on a lab cart that is carrying some masses, the acceleration is 3.0 m/s2. If the force is tripled and exerted on the cart with a total mass of one half the original, then the acceleration will be



    2. f=ma



    3. do i just times 3.0m/s2 by three? or do i have to somehow find more variables then solve and equation?
     
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  3. Jul 16, 2009 #2

    PhanthomJay

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    Re: Force

    No, that won't cut it. If the letter algebra bothers you, try putting some numbers to the problem. Like assume the original mass is 10 kg, in which case the net force is (10)(3) = 30 N. Now what happens to the accelertaion when the force is tripled (90 N) and the mass is halved (5 kg)?
     
  4. Jul 16, 2009 #3
    Re: Force

    when i divide the Force by the Mass, ur example gives me 18. which makes sense. But the question im trying to figure out is multiple choice, so it could be any of the answers if i pick the right number..
     
  5. Jul 16, 2009 #4

    PhanthomJay

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    Re: Force

    What are the choices? Supposing you assumed the original mass was 20 kg, what does the acceleration work out to when the force is tripled and the mass is halved?
     
  6. Jul 16, 2009 #5
    Re: Force

    the choices are,

    1.5m/s2
    6.0m/s2
    9.0m/s2
    18m/s2

    and if i assumed that it was 20, then 6 would be the right answer. 6 would also be right if i assumed the mass was 15
     
  7. Jul 16, 2009 #6

    PhanthomJay

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    Re: Force

    I think I may be confusing you instead of helping you, sorry. 6m/s^2 is not the right answer. 6 is the factor by which the accelertaion increases.
    F=ma, or a=F/m. When F is tripled and m is halved, then the new accelertaion is 3F/.5m = 6F/m = 6a. Does that make it simpler? Or harder? Solve for the new a = ?????m/s^2
     
  8. Jul 16, 2009 #7
    Re: Force

    so by multiplying six by the original acceleration (3), that gives me 18. would that be right then?
     
  9. Jul 16, 2009 #8

    PhanthomJay

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    Re: Force

    I would think so, but please convince me and yourself that you thoroughly understand why that is the correct answer. Thanks.
     
  10. Jul 16, 2009 #9
    Re: Force

    thank you so much.

    and yes i do understand. since three times the force, divided by 0.5 is 6, then that is the coefficient for your original acceleration. ( times those two together)
     
  11. Jul 16, 2009 #10

    PhanthomJay

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    Re: Force

    OK, great.
     
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