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Force and Acceleration/Force of Friction and Air Resistance. 5 Basic Questions.

  1. Oct 25, 2009 #1
    Q1: Assuming friction force to be negligibly small, what will be the acceleration in m/s^2 of a 45kg skater when propelled by a force of 250N?
    My attempt
    Data:

    F-250N
    m-45kg
    a-m/s^2

    Formula:
    F=ma
    a=F/m
    a=250N/45kg
    a=5.555
    a=5.6m/s^2
    Q2: a 0.5kg mass on a horizontal frictional free table is accelerated by a force from a string passing over a pulley and attached to a mass of 1.5kg hanging vertically.

    My attempt
    Data:

    m-0.5kg
    hanging m-1.5kg

    formula:
    F=F1+F2
    F=0.5kg+1.5kg=2.0kg
    F=ma
    F=2.0kg(9.80m/s^2)
    F=19.6 rounded to 20N

    My attempt at B)
    Data:

    F=19.6N
    m=2kg
    a=?m/s^2?

    Formula:
    F=ma
    a=F/m
    a=19.6N/2kg
    a=9.8 rounded to 10m/s^2
    Q3: A 1200kg truck is moving at 108km/h and stops within a distance of 91m
    a)What is the acceleration?

    My attempt
    Data:

    m-1200kg
    Vi-108km/h--->converted to 30m/s
    Vf-0
    s-91m

    Formula:
    2as=Vf^2+Vi^2
    a=Vi^2/2s
    a=(30m/s)^2/2(91m)
    a=900m^2/s^2/182m
    a=-4.9m/s^2(negative because it decelerated)

    My attempt at part B)What is the coefficient of friction?
    Formula:

    Ff=uFn
    u=Ff/Fn
    u=4.9m/s^2/1200kg
    u=0.00408
    Q4: you are pulling a 3.0slug crate and a sliding friction is known to be 55lb. what is the acceleration o the crate when the pulling force is?
    a)55lb, b)75lb and c)105lb

    My attempt at a)
    Date:

    m: 3.0Slug
    F:55lb
    a:?m/s^2?

    Formula:
    F=ma
    a=F/m
    a=55lb/3.0slug
    a=18ft/s^2

    My attempt at part b)
    Data:

    m: 3.0slug
    F: 75lb
    a: ??

    formula:
    F=ma
    a=F/m
    a=75lb/3.0slug
    a=25ft/s^2

    My attempt at part c)
    Data:

    m: 3.0slug
    F:105lb
    a-??

    Formula:
    F=ma
    a=F/m
    a=105lb/3.0slug
    a=35ft/s^2
    Q5: During a training exercise, Brian who weighs 75kg skydives and parachutes from a stationary helicopter. What is the acceleration when the air drag is a)500N, b)1500N, and c)Equal to his weight

    My attempt at a)
    Data:

    Fw:75kg
    g:-9.80m/s^2
    m: 500N
    a=?

    Formula:
    F=ma
    a=F/m
    a=500N/75kg
    a=6.66m/s^2

    My attempt at b)
    Formula:

    a=F/m
    a=1500N/75kg
    a=20m/s^2

    My attempt at c)
    Formula

    a=f/m
    a=75N/75kg
    a=1 m/s^2
    first of all I want to thank anyone that gives their time and effort to help me do this homework that is due very soon. I don’t think I understand it very well and I am start to get really worried. Secondly, if you notice any problems, can you please show me how to do it? Id appreciate any help I can get. Thanks a ton people.
     
  2. jcsd
  3. Oct 26, 2009 #2
    Q1: ok

    Q2; you didn't copy question A. F=19.6 N is wrong however (and F=2.0kg is wronger)
    Let T be the tension in the rope, and then find the net force on both blocks
    from gravity and this unknown tension. Then use F=ma for both blocks, and the fact
    that a is the same for both blocks to find T.

    Q3-a: The answer is OK. You should have been using 2as=Vf^2-Vi^2 and then a would have
    become negative by itself. The sign of a also depends on what direction is chosen as
    the positive direction. Here it's the direction that the car drives, but that isn't always the case.
    Q3-b: 4.9 m/s^2 isn't a force and neither is 1200 kg. What are the forces?
    Q4: I don't do slugs, but you need to find the net force, wich includes the friction force that is opposite to the pulling force.
    Q5: You need to find the net force here as well. In physics weight is a force, the force of gravity, and it is equal to the mass times the acceleration of gravity mg with g = 9.8 m/s^2
    so the weight of the person is 735N and not 75 kg.
     
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