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Acceleartion and Force again !

  1. Oct 12, 2007 #1
    #1
    Ricky is on a rocket ship far from any planet or sun. The 900kg rocket fires its rockets for 3.5s sec and goes from 0m/s to 50 m/s.(a)What acceleration does the rocket undergo ?(b) If 160 kg of gas is expelled from the rocket. what is the gasses acceleration ?
    ------------------
    vi = 50 m/s vf= 0 m/s
    t = 3.5 s
    m = 900kg
    a) vf = vi +at
    -50 = 3.5a
    a) a = 14.3 m/s
    b) m= 740 kg
    stuck right here ?

    #2
    Award drives a 5000 kg car cross a bridge that inclines up at 10' above the horizontal. While going cross the bridge with an acceleration of 4 m/s2 the car encounters a m_rolling of .05. With what force does the car push on the ground in order to move forward ?
    ----------------
    F (per) = 49000 cos 10 = 48255.6 N
    F (par) = 8508.8 N
    F_net = 5000 (4) = 20000 N
    F_fricton = .05 (48255.6) = 2412.8 N
    F_net - F (par) - F_friction = F_apply
    F_apply = 9078.4 N
    Is this right /

    #3
    A box is sitting on the level ground U (static) = .2and it has a mass of 100 kg. A person wants to pall the box 25 m cross the room with a rope, but can only exert a force of 310 N. That person pulls the rope at 32* and fins the U (sliding) = .12. Does the box move and if so how much time does it take for that person to move across the room.
    F_y = 0
    F_y = n + 310sin32 – 980
    Normal force = 815.7 N
    F_friction = .2(815.7) = 163.14 N
    F_x = 310cos32 – 163.14 = 99.7 N
    Got stuck right here ?
     
  2. jcsd
  3. Oct 12, 2007 #2

    learningphysics

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    Use newton's third law. What is the force acting on the rocket. The same force acts on the gas.

    I think it should be: Fnet = Fapply - Fpar - Ffriction. then solve for Fapply.

    The applied force: 310cos32 is greater than static frcition 163.14... so that means the box moves.

    Now, you need to find the net force in the x-direction using kinetic friction. Then get acceleration and then you can get time using kinematics.
     
  4. Oct 13, 2007 #3
    #1
    Is that hints for part b

    #2
    F_apply = 30921.6 N

    #3
    Is that mean I was going on the right track right ?
    F_apply = 262.9 N
    a) the box moves
    b)
    F_friction = .12 (815.7) = 97.9 N
    F_x = 310cos32 - 97.9 = 164.9 N
    F = ma
    164.9 = 100a
    a = 1.65 m/s2
    25 = (1/2)(1.65)t2
    t = 5.5 s

    Am i doing right ?
     
  5. Oct 13, 2007 #4

    learningphysics

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    Yup everything looks right to me. good job! so now, just the first problem... yes, that hint was for part b).

    Part a) looks like you switched vf and vi. vf = 50. vi = 0. but the acceleration you got is right. 14.3m/s^2 (watch the units).
     
  6. Oct 13, 2007 #5
    Problem 1
    a) how come it is so easy ?
    b) got a little confuse, there only force of gravity is act on it.
     
  7. Oct 14, 2007 #6

    learningphysics

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    No gravity since it is away from any planet...

    The gas pushes the rocketship forward. The rocketship pushes the gas backwards...

    What is the net force acting on the rocket? You've got the mass and acceleration...

    So what is net force acting on the gas? Given the mass of the gas, calculate the acceleration.
     
  8. Oct 14, 2007 #7
    F_net (acting on the rocket) = 9000 (14.3) = 128700 N
    F_net (rocket) = F_net (gas) (From your hints if this two force were equal, then how come the rocket laugh forward)
    128700 = 740 a
    a = 173.9 m/s2
    Is this right ?
     
  9. Oct 14, 2007 #8

    learningphysics

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    Assuming the 900kg is the mass of the rocket + the mass of the gas...

    The mass of the rocket alone is 740kg.

    The mass of the gas is 160kg.

    Force on the rocket = ma = 740kg*14.2857m/s^2 = 10571.43N

    Force on the gas = 160kg*a
    10571.43N = 160a
    a = 66.07m/s^2

    so acceleration of the gas is 66.07m/s^2
     
  10. Oct 14, 2007 #9
    OK, it was tricky !
    And i was confused, how come the part a was so easy. Usually we use F=ma to find the acceleration when the problem involved with mass.
     
  11. Oct 14, 2007 #10

    learningphysics

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    Part a) is just kinematics. Usually yeah, they give the force and then we get the acceleration from that.

    Part b) can also be done with "conservation of momentum" if you've studied momentum.

    Net momentum of the system initially is 0. (the rocket and gas is at 0m/s... momentum is 0*900kg = 0)

    Final momentum is (mass of rocket)*(velocity of rocket) + (mass of gas)*(velocity of gas)

    initial momentum = final momentum

    0 = (mass of rocket)*(velocity of rocket) + (mass of gas)*(velocity of gas)
    0 = 740*50 + 160*vgas

    vgas = -231.25 (the gas is in the backwards direction)

    then acceleration of gas = (-231.25 - 0)/3.5 = -66.07m/s^2

    or in other words 66.07m/s^2 in the backwards direction.
     
  12. Oct 14, 2007 #11
    Not cover momentum yet. but that was great. Thank a lot learning physics.
     
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