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Question about Force

  1. Nov 25, 2007 #1
    Since I have a textbook without that ans for a certain section, that's why I wanna check did I do these questions in the right way. Please tell me the right answers and show me the steps if I got them wrong. Thank you :wink:
    This is the question:

    The instrument attached to a weather balloon have a mass of 5.0kg.

    a. The balloon is released and exerts an upward force of 98N on the instruments,What is the acceleration of the balloon and instruments?

    I got 20m/s^2

    b. After the balloon has accelerated for 10 s, the instruments are released. What is the velocity of the instrument at the moment of their release?

    I got 196 m/s

    c. What net force acts on the instruments after their release?

    I got 147N

    d. When does the direction of their velocity first become downward?

    When the instrument stopped going upward and start falling.

    Once again, thank you so much for helping me :!!)
  2. jcsd
  3. Nov 25, 2007 #2
    a. The 5kg mass has a weight of about 49N (W = mg), this is in a direction opposite to the lift of the balloon. Since they're in opposite directions, we can subtract them to find the resultant forces. 98 - 49 = 49 Newtons up.
    F = ma, so the force we just worked out, 49 = 5x, x ~ 10. The acceleration is 10ms^-2.

    b.) vf = vi + at, since the balloon started at rest, vi = 0. So vf = at, 9.8x10 = 98ms^-1.

    c.) The net force is ONLY the gravity, as the balloon has stopped applying a force on the instruments. The force of gravity is about 49N (9.8 x 5). Remember, the force on an object doesn't matter what the speed is. An object could be going close to the speed of light, and have no force acting on it. All it means is that it's going constant velocity.

    d.) The object when it's released still has velocity, which you worked out in question b.). So only when that velocity is gone from gravity, will it start to fall. It's quite hard to visualize. But just imagine throwing a ball from a car on the highway. It would bounce along in the direction that the car is moving.

    So for this:
    vf = vi + at
    vf = 0, as we're finding the point when the instruments momentarily stop in the air.
    vi = 100.
    at = -9.8t (it's negative because we chose up as being positive, and down as being negative).

    0 = 100 + -9.8t
    9.8t = 100
    100/9.8 = t = 10s.

    About 10 seconds after the object has been dropped, it will start to fall down.
  4. Nov 25, 2007 #3


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