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  1. MHB Craftsman

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    #1


    Calculate the velocity of the brick at the bottom of the inclined plane.

    State the assumption you made for your calculation in part above.

    & Sketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)

    Workings:

    I am unable to think on the first two but I think the displacement and time graph should be something like ,



    Many Thanks

  2. Pessimist Singularitarian
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    #2
    To calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.

    So, if we equate the initial potential energy to the final kinetic energy, we have:

    $ \displaystyle mgh=\frac{1}{2}mv^2$

    What do you get upon solving for $v$?

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    #3 Thread Author
    Quote Originally Posted by MarkFL View Post
    To calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.

    So, if we equate the initial potential energy to the final kinetic energy, we have:

    $ \displaystyle mgh=\frac{1}{2}mv^2$

    What do you get upon solving for $v$?
    Thank you very much MarkFL

    $ \displaystyle mgh=\frac{1}{2}mv^2$

    $ \displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}mv^2$

    $ \displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}*2kg * v^2$

    $ \displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}*2kg * v^2$

    $ \displaystyle 100 J= v^2$

    $ \displaystyle 10 J= v$

    Correct ?

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    #4
    I would solve the equation first, and then plug in the numbers:

    $ \displaystyle mgh=\frac{1}{2}mv^2$

    $ \displaystyle v=\sqrt{2gh}$

    Okay, at this point we can plug in:

    $ \displaystyle g\approx9.81\,\frac{\text{m}}{\text{s}^2},\,h=5\text{ m}$

    $ \displaystyle v\approx\sqrt{2\left(9.81\,\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ m}\right)}=3\sqrt{\frac{109}{10}}\,\frac{\text{m}}{\text{s}}\approx9.9045\,\frac{\text{m}}{\text{s}}$

  5. MHB Craftsman

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    #5 Thread Author
    Quote Originally Posted by MarkFL View Post
    I would solve the equation first, and then plug in the numbers:

    $ \displaystyle mgh=\frac{1}{2}mv^2$

    $ \displaystyle v=\sqrt{2gh}$

    Okay, at this point we can plug in:

    $ \displaystyle g\approx9.81\,\frac{\text{m}}{\text{s}^2},\,h=5\text{ m}$

    $ \displaystyle v\approx\sqrt{2\left(9.81\,\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ m}\right)}=3\sqrt{\frac{109}{10}}\,\frac{\text{m}}{\text{s}}\approx9.9045\,\frac{\text{m}}{\text{s}}$
    Thank you very much

    speaking about the graph

    Quote Originally Posted by mathlearn View Post

    Sketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)


    Is the above drawn graph correct?
    Last edited by mathlearn; November 2nd, 2016 at 04:51.

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    #6
    For the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?

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    #7
    Quote Originally Posted by mathlearn View Post
    speaking about the graph

    Is the above drawn graph correct?
    Hey mathlearn!

    Here's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):

    On the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.
    As a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).
    Note that $v=\int_0^t a\,dt$.

    However, we're asked for the displacement, which we can call $d$.
    We have that $d=\int_0^t v\,dt$.
    What would the displacement graph look like?

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    #8
    Quote Originally Posted by mathlearn View Post
    $ \displaystyle 100 J= v^2$

    $ \displaystyle 10 J= v$
    Recheck your units, speed is not measured in J. (And the square root of a J is not J!)

    -Dan

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    #9 Thread Author
    Quote Originally Posted by MarkFL View Post
    For the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?


    This kind of a curve with constant acceleration and with deceleration.

    Quote Originally Posted by I like Serena View Post
    Hey mathlearn!

    Here's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):

    On the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.
    As a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).
    Note that $v=\int_0^t a\,dt$.

    However, we're asked for the displacement, which we can call $d$.
    We have that $d=\int_0^t v\,dt$.
    What would the displacement graph look like?
    The displacement time graph would look like the above graph

    Quote Originally Posted by topsquark View Post
    Recheck your units, speed is not measured in J. (And the square root of a J is not J!)

    -Dan
    Thanks for the catch
    Last edited by mathlearn; November 2nd, 2016 at 23:46.

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    #10
    Quote Originally Posted by mathlearn View Post
    This kind of a curve with constant acceleration and with deceleration...
    Why do you have the brick returning to zero displacement?

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