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Net Force Questions

  1. Oct 19, 2005 #1
    1. A 1.2 x 10^4 kg truck is travelling south at 22 m/s.
    a) What net force is required to bring the truck to a stop in 330 m?
    b) What is the cause of this net force?
    Answer: a) 8.0 x 10^3 N [N]

    2. Each of the four wheels of a car pushes on the road with a force of 4.0 x 10^3 N [down]. The driving force on the car is 8.0 x 10^3 N [W]. The frictional resistance on the car is 6.0 x 10^3 N [E]. Calculate the following:
    a) the mass of the car
    b) the car's acceleration
    Answer: a) 1.6 x 10^3 kg b) 1.2 m/s^2 [W]

    I have the answers to these two questions, but I really need help on how to understand and do it.

    So far...
    1. a) v = d/t
    t = d/v
    t = (0.33 km)/(22 m/s )
    t = 0.015 s

    a = v/t
    a = (22 m/s )/(0.015 s)
    a = 1466.7 m/s^2 ?? <====== stopped here because this makes no sense..
     
    Last edited: Oct 19, 2005
  2. jcsd
  3. Oct 19, 2005 #2
    the time makes no sense... you are right :p

    here is your problem

    t = (0.33 km)/(22 m/s )

    the distance has to be in meters...
     
  4. Oct 19, 2005 #3
    If you've covered energy then that would be the best way to solve the problem. If not then the following equation should look familiar:

    [itex] v_f^2 = v_i^2 + 2ad [/itex] which is what you should use to tackle this problem.
     
  5. Oct 19, 2005 #4
    whozum... is that to find the acceleration?

    (22 m/s )^2 = (0 m/s )^2 + 2a(330 m)
    a = 0.73 m/s^2

    f = ma
    f = 8760 N <----- something's still wrong
     
    Last edited: Oct 19, 2005
  6. Oct 19, 2005 #5
    if you want the truck to stop acceleration has to be negative...
     
  7. Oct 19, 2005 #6
    that still doesn't work though because the net force comes up to -8670 N .

    i think i'm totally missing out on something here...
     
  8. Oct 19, 2005 #7

    The acceleration is not south but north, -8670 N = 8670N [N]. However I don't know why they have 8.0 x 10^3N as the answer, your answer of 8670 is alomst correct.

    By the way the correct solution is

    [itex] v_f^2 = v_i^2 + 2ad [/itex] you had the velocities mixed up.

    [itex] (0)^2 = (22)^2 + 2a(330m) [/itex]

    [itex] a = \frac{22^2}{-660} = -0.733333 [/tex]

    [itex] F = ma = m(-0.7333333) = 8800N = 8.8 \times 10^3 N [N] [/itex]
     
    Last edited: Oct 19, 2005
  9. Oct 19, 2005 #8
    okay thankssss... and for number 2... i really dont know how itll work cause mass is equal to net force over acceleration, but all i have is net force... no acceleration.... so how would that work out?
     
  10. Oct 19, 2005 #9
    The force from the 4 tires is the weight of the car (normal force). You can find the mass from there.
     
  11. Oct 19, 2005 #10
    does that mean that the answer is wrong again? because i came up with 16000 N which is 1.6 x 10^4 and they say its 1.6 x 10^3 kg.

    and does 1 N equal to 1 kg.... im not very familiar with that.
    but 1 N actually equals to 1 kg x m/s^2, though right?
     
    Last edited: Oct 19, 2005
  12. Oct 19, 2005 #11
    you answered your own question a newton is a kg * m/s^2
     
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