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Questions about force

  1. Nov 19, 2009 #1
    Hi there, I have a physics test tomorrow and I've been trying to review and practice some problems, but I'm not sure I'm understanding it. When the teacher goes through the questions it makes sense, but I'm not always sure where to start. I have a couple questions that I don't understand, even if I could understand one of them it would be good.

    1. The problem statement, all variables and given/known data
    What is the average force exerted by a shot-putter on a 7.0 kg shot if the shot is moved through a distance of 2.8m and is released with a speed of 13 m/s?

    So I know
    m=7.0 kg
    v=13 m/s
    d=2.8m
    g=9.8 N/kg or 9.8 m/s
    and I'm trying to find force.


    2. Relevant equations
    a=v/t
    F=ma
    Not really sure which equations to use.


    3. The attempt at a solution

    I'm not sure if the distance is even relevant, does it help with the problem at all?
    The first thing I thought to do was find the acceleration using a=v/t and then using F=ma, but I don't know the time and wouldn't the acceleration be 0? So if a=o, then F=ma would also be 0, and that doesn't make sense. Overall I'm just not too sure where I would start if I was given a question like this on the test.
    The book says the answer is 2.1 x 10N.

    1. The problem statement, all variables and given/known data
    A 6500 kg helicopter accelerates upwards at 0.60 m/s while lifting a 1200 kg car. a) What is the lift force exerted by the air on the rotors? b) What is the tension in the cable (ignore its mass) that connects the car to the helicopter?

    total mass=7700kg
    a=0.60m

    3. The attempt at a solution

    So for a) I tried to do F=ma
    F= (7700kg) (0.60m) = 4620 N
    But the book says the answer is 8.01 x 10[tex]^{}4[/tex]N.
    and for b) i honestly have no idea where to start, but the answer is 1.25 x 10[tex]^{}4[/tex]N.

    I would really appreciate it if anyone could help me with these problems, thank you very much!
     
  2. jcsd
  3. Nov 19, 2009 #2
  4. Nov 19, 2009 #3
    try using v^2 = v0^2 + 2a(x0 - x) to solve for a
     
  5. Nov 19, 2009 #4
    First Question:
    Work = Force x Distance ( W = Fd )

    The amount of work done will result in an equivalent change of Kinetic Energy:

    Ek = mv2/2​

    Knowing the mass and velocity, you can calculate the kinetic energy, and then force.

    ETA:
    For your second question, don't forget about the acceleration of gravity (g).
     
    Last edited: Nov 19, 2009
  6. Nov 19, 2009 #5
    do you really need these for this question? I thought that these were just standard f = ma questions
     
  7. Nov 19, 2009 #6
    are you sure that the answer for B is 833.04 newtons? it doesn't seem right..
     
  8. Nov 19, 2009 #7
    No, you're right. I'm just used to doing it that way.

    v^2 = 2ax : solve for a

    For the second question:
    m = 6500kg + 1200kg
    a = 9.8m/s^2 + .6 m/s^2

    I think the given answer is 8.01x104 N.
     
  9. Nov 19, 2009 #8
    Ohh, okay right.. That makes more sense, thanks

    I think it should be 0 = Vo^2 + 2a([tex]\Delta[/tex]x) ? it says that it is 13 m/s when released, right?
     
  10. Nov 19, 2009 #9
    I notice that you probably want some explanations, since F= (7700kg) (0.60m) = 4620 N as an answer for part B wouldn't really make any sense. You know that after all the forces are calculated, the helicopter moves up with an acceleration of 0.60 m/s^2. So, we know the forces pulling down on the helicopter and the net force - all we need to do now is equate the pulling down forces and some mysterious force (to be known as the rotor air lift force) to be equal to the net force. So if we have forces pulling down on the helicopter, there must exist a force making it go up, since the net force makes it go up
     
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