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Questions on motion

  1. Jun 9, 2007 #1
    1. Explain why the weight of a body changes if it is taken form the equator to one of the poles
    2. With Reference to Newton’s Laws of Motion, determine the force that a man of mass 75.0kg exerts on the bottom of an elevator when it is
    a. Ascending with a constant velocity of 2.5m/s
    b. Descending with an acceleration of 4.0m/s2
    3. A 100kg pile driver is suspended 12.0m above the ground where it is eventually released and falls, driving a pile into the ground
    a. What is the gravitational potential energy before it is released and ½ way of its fall
    b. Why the difference in gravitational potential energy above




    [bThe attempt at a solution:[/b]
    1. I will have to research question one, but any help would be appreciated.
    2. Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. Newton's second law can be summarized by the formula:

    f=ma
    a) When ascending: f=75kg*2.5m/s^2 = 187.5 N
    b) When descending: f=75kg*4m/s^2 = 300 N

    I have a feeling that I am wrong because I didn't consider the motion of the elevator...

    G.P.E=mgh= 100kg*10m/s^2*12m = 12000 J
    G.P.E at half the distance= 1/2*12000J= 6000 J

    There is a difference of the two G.P.E's above because G.P.E is defined by the formula: mass(m)*gravitational acceleration(g)*height(h). At half-way during the fall, the distance is also half so as a result the value of mgh would now be mgh/2 as opposed to the GPE just before the fall (mgh), and hence the difference in the two.

    I think I have done a fair effort and it would be even better if I am corrected and guided, thank you.
     
  2. jcsd
  3. Jun 9, 2007 #2

    cepheid

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    Hint: consider Newton's Law of Gravitation


    Part (a) is wrong. You mistook the 2.5 m/s velocity for a 2.5 m/s^2 acceleration. There is no acceleration. The elevator is moving up at a constant velocity. As a result, the net force on the system (elevator + man) = 0. Does this help you to answer the question? Look at it another way...when you are in an elevator that is not accelerating, you don't feel any heavier or lighter than normal, do you?

    Part (b) doesn't look quite right either. I'd recommend drawing a free body diagram for the elevator + man. There are only two forces:

    1. The weight of the (elevator + man) downward
    2. The force pulling the elevator upward (e.g. from a cable).

    Clearly, since the elevator is accelerating downward, the second force is slightly smaller than the first by an amount equal to (total mass)*a. To put it another way, there is a net downward force on the (elevator + man).

    I haven't checked your math, but it's pretty trivial (just mgh) so I'm sure you got it right. As for the explanation, you're right. Appealing to the mathematical definition of potential energy is the most succint and therefore minimalist answer. However, to round out your explanation, I think it would be a good idea to explain what happened to the initial potential energy after having fallen halfway, i.e. where did it go? That way, you'd also be explaining why the potential energy changes with height and is given by mgh.



    Agreed
     
  4. Jun 11, 2007 #3
    So could it be:

    a) Weight of man (assuming g=10m/s^2)= 75*10=750N

    Net force= force * acceleration
    However, since there is a constant velocity, there would be no acceleration. Therefore: Net force= 750*0= 0N

    Since there is no unbalanced force resulting in an acceleration, the force that the man exerts on the elevator is: 750 Newtons.

    b) Net downward force= 75*4 = 300 N
    Weight of man= 75*10= 750 N
    Therefore, actual force exerted by man= 750 - 300 = 450 N

    Am I correct? I have a gut feeling that the accleration downwards is a negative one...
     
  5. Jun 11, 2007 #4

    cepheid

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    Although your result may be right, I'm not 100% sure about how you arrived at it. This is the method that makes sense to me personally. Since you know the acceleration of the elevator, you know the acceleration of the man. So you can draw a FBD for the man only. There are two forces acting on him:

    1. weight downward (call this w)

    2. force pushing up on his feet (call this N).

    We get for the net force F:

    F = N + w

    We know that F is negative (the accelerating of the elevator is downward), which suggests that N < w (the floor is not pushing up on the dude as much as his weight...as a result, he feels lighter. If the elevator were accelerating downward at g, the floor wouldn't push up on him at all. He'd be in free fall and would feel weightless). In this case, he just feels lighter.

    plug the numbers in (using a "downward is negative" convention):

    75*(-4) = N + 75*(-10)

    N = 75*(-4) - 75*(-10) = -300 N - (-750 N) = 450 N

    The floor pushes up on the man with this force (*less* than his weight).
     
    Last edited: Jun 11, 2007
  6. Jun 12, 2007 #5
    Thank you very much.
     
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