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Hold 發問於 科學及數學其他 - 科學 · 1 十年前

物理 Air Track 的實驗問題

我做了一個 Air Track 實驗, 是把一個 glider 放在無 friction 的 air track 上, 然後用一些 blocks 把 air track 頂高. 跟著把 Glider 由最高點放下. Since blocks 把 air track 頂高了, 所以 acceleration a of an object due to gravity down the incline of angle theta 是 a = g sin ( theta )

我怕說得不 clear, 這是我的 lab :

http://www.sci.ccny.cuny.edu/physics/LabMan/airtra...

我的問題是 Part B 和 Analysis

我想 momentum is conserve, but don't know if it is right or wrong.

Part B (2), (3) 不知怎樣用, is that right to use the (K-K')/K for question 2?

Analysis (1) : 我估計 gravity and Normal Force are the only two forces acting on it

But How do I find the net force and apply Newton's Second Law ...

(2), (3), (4) ...

(5) I know is to use my gravity I found to subtract the standard one ...

Thank you so much for helping. I will study more about collision while I am waiting for someone to answer.

更新:

For Analysis (1), it says find the net force and apply Newton's Second Law to determine algebraic relation between the acceleration a, g, and angle theta of the track

For the net force, does that mean I have to add KE + PE = KE' + PE' since there is no external force?

3 個解答

評分
  • 天同
    Lv 7
    1 十年前
    最愛解答

    Q:Part B (2), (3) 不知怎樣用, is that right to use the (K-K')/K for question 2?

    It is already given in your lab manual (the first paragraph on p.2) that the fractional loss of kinetic energy equals to [1 - (D'/D)]. By measureing D' and D, you should be able to find the fractional energy poss.

    The prove is rather simple. The kinetic energy (K) before collision at the spring = potential energy (PE) at the top of the track. Thus,

    K = mg.D.sin(theta)

    Similarly, kinetic energy (K') after collision at the spring = PE at the highest point after collision

    K' = mg.D'.sin(theta)

    Fractional loss of kinetic energy

    = (K-K')/K = [(D-D')/D] = 1 - D'/D

    Q: Analysis (1) : 我估計 gravity and Normal Force are the only two forces acting on it. But How do I find the net force and apply Newton's Second Law ...

    There are only two forces acting on the glider. The weight of the glider, mg, which acts vertically downward. The normal reaction given to the glider, which acts perpendicular to the inclined plane.

    The resultant of these two forces is in the direction acting downward along the inclined plane.

    By force addidtion, you could find that the resultant force is given by mg.sin(theta).

    The acceleration a of the glider is thus related to the resultant force by

    ma = mg.sin(theta)

    5) I know is to use my gravity I found to subtract the standard one ...

    [delta g] is the measurmentl uncertainty. The equation just compare whether the deviation from the standard g value is within the measurment uncertainty. If not, there would be mistake in the experiment.

    Just follow the method describes in para.4 to determine [delta g].

    2010-03-23 08:58:55 補充:

    In response to your "opinion", there is indeed a net force, which is the resultant of the glider's weight and the normal reaction. This net force acts downward along the inclined plane. Thus, ma = mg.sin(theta), and you finally got a = g.sin(theta)

  • 1 十年前

    For Analysis (1), it says find the net force and apply Newton's Second Law to determine algebraic relation between the acceleration a, g, and angle theta of the track

    For the net force, does that mean I have to add KE + PE = KE' + PE' since there is no external force?

  • 匿名
    1 十年前

    Instead of giving you all the calculation, it is more important to unders the principles. Here are some tips when you encounter this kind of problems:

    Conservation of momentum is applied only when there is NO EXTERNAL force on a system. If you consider a falling object alone, there is an external force (namely gravitational force) on it,so the momentum of a falling object is NOT constant/conserved. The momentum of a falling object actually increases as the the object is falling because the speed is increasing.

    To apply f=ma, the direction of the force must be the same as the direction of the acceleration. If you have gravity acting vertically downward while the object is moving at an angle from the vertical line, you will have to break the gravitational force in the 2 components -- one component in the same direction as the movement of the object and the other component acting at 90 degrees to the direction of the movement of the object. And in f=ma, the force is the component of gravitational force acting in the same direction as the movement of the object.

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