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Newton's laws

  1. Sep 10, 2006 #1
    "a person drives their 1452 kg car along a straight road at a constant velocity of 30 m/s east. their brakes suddenly give out. they put the car in neutral and let is coast for 25 seconds. the air drag decelerates the car to a velocity of 25 m/s east. (assume a frictionless surface)."

    b) assume the average acceleration while the car is decelerating

    so it's just (25 m/s - 30 m/s)/25 s = -0.2 m/s^2?

    c) determine the average force of air against the car
    f = ma
    = (1452 kg)(-0.2)
    = 290.4 N

    d) after coasting for 25s. they pull their car handbrake to slow the car to a stop. this take 3 seconds. what is the force applied by the handbrake? assume that the force exerted by the air remains constant and is equal to the forced determined in part (c).
    so i assume i find acceleration by (0 - 25)/time

    but for the time is it (25s-3s) or (3s - 25s)?

    and then i just the acceleration times the mass of car = handbrake force?

    any help will be appreciated!

    ~Amy
     
  2. jcsd
  3. Sep 10, 2006 #2
    You have to consider only the last 3 seconds during which the car is moving.

    acceleration times the mass of car = handbrake force?
    That's right.
     
  4. Sep 11, 2006 #3
    thanks neutrino.

    for d) i got -8.33 m/s^2 for the acceleration abd -12100N for the force of the handbrake.

    in e) it asks "what is the total displacement from the time the brakes give out to the time the car stops"

    so for this i used the formula:
    d = (0.5)at^2 + v1t
    = (o.5)(-8.33)(3^2) + (25m/s)(25s)
    = 587.52m

    can someone tell me if this is accurate?

    ~Amy
     
  5. Sep 11, 2006 #4
    another question i want to check:

    "peter and john are playing tug of war on a frictionless surface. peter weighs 539N and john weights 393N. john accelerates towards peter at 3 m/s^2. "

    a) what is the magnitude of the force that peter exerts on john?

    539N times 3 m/s^2
    = 1617N?

    any help is appreciated!
    ~Amy
     
  6. Sep 12, 2006 #5
    for this one:

    "peter and john are playing tug of war on a frictionless surface. peter weighs 539N and john weights 393N. john accelerates towards peter at 3 m/s^2. "

    a) what is the magnitude of the force that peter exerts on john?

    first i calculated john and peters mass by diving this newton weight by the force of gravity (9.8 m/s^2). so peter is 55 kg, and john is 40 kg.

    then for part a) i took peters mass (55 kg) times (-3 m/s^2)
    = -165N.

    not sure if this is accurate. help?

    ~Amy
     
  7. Sep 12, 2006 #6

    Doc Al

    User Avatar

    Staff: Mentor

    No. Don't forget that the air is still exerting its force.

    No, you found the net force.

    Divide the motion into two segments and find the displacement in each:
    (1) air drag only (coasting)
    (2) air drag + hand brake
     
  8. Sep 12, 2006 #7

    Doc Al

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    Staff: Mentor

    OK.

    No. To find the force on John, multiply John's mass times John's acceleration.
     
  9. Sep 12, 2006 #8
    so i'd be -121000N - -290.4N?

    will do (and will post my results later)

    :blushing: thanks!
    so it'd be 120N

    what then how would i find the magnitude of the force that john exerts on peter? would it be -120N?

    thank you

    ~Amy
     
  10. Sep 12, 2006 #9

    Doc Al

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    Staff: Mentor

    Yes.



    Right.

    Yes. Newton's 3rd law should tell you that they exert equal and opposite force on each other. (But realize that magnitudes are always positive. :wink: )
     
  11. Sep 12, 2006 #10
    ok. for (1) i got (0.5)(30m/s + 25 m/s)(25s)
    = 687.5 m

    for (2) i got (0.5)(-8.33 m/s^2)(3^2)
    = -37.485

    so 687.5 + - 37.485
    = 650.02 m

    :biggrin:

    more on the john/peter saga..
    so first of all she weighs 30kg

    and to find the c^2 it's (45^2) + (25^2)
    = 51.48 N

    and to find the tan angle is 26 degree north/west.

    so to find sarah's acceleration i go 51.48N/30 kg?

    ~Amy
     
  12. Sep 12, 2006 #11

    Doc Al

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    Staff: Mentor

    Good.
    Nope. Sanity check: It slows down, but doesn't go backwards. :eek:

    Use the full equation:
    [tex]d = v_0 t + (1/2) a t^2[/tex]
    (don't forget the initial velocity!)

    You can also use the exact same equation you used for segment (1). :wink:


    Good.

    Check this one. Her acceleration should be north-east.

    Good.
     
  13. Sep 12, 2006 #12
    (i knew the negative displacement sounded a bit suspicious, lol)

    ok so (25m/s)(3s) + (0.5)(-8.333)(3^2)
    = 37.515

    650.02+37.515
    = 687.54 m :biggrin:

    thanks for the help!

    ~Amy
     
  14. Sep 12, 2006 #13

    Doc Al

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    Staff: Mentor

    Good. But you can also use the same equation that you used for segment (1): d = (0.5)(25 m/s + 0 m/s)(3 s)


    I think you copied the answer from (1) wrong.

    My pleasure.
     
  15. Sep 12, 2006 #14
    thanks. i just noticed that too. so the total displacement should be 725.02 m east :smile:

    ~Amy
     
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