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Newtonian Mechanics, Motion on Inclines, Work Energy Theorem

  1. Apr 18, 2005 #1
    Let's start with the very base of Newtonian mechanics. It works like this : Suppose we work in two dimensions denoted by a x-axis and an y-axis. You can work in as many dimensions as you want because all you have to do is add a unit vector to the formula's, as you will see.

    Starting from the acceleration ,one can calculate the velocity and position by using integrals at any time : r_0 is initial position at t=0, v_0 is initial velocity

    [tex]\vec{F} = F_x \vec{e_x} + F_y \vec{e_y} = m(a_x \vec{e_x} + a_y \vec{e_y})[/tex]

    Where the e_x and e_y denoted the x and y-direction (ie the unit vectors)

    Now, integrating will yield
    [tex]\vec{v} = \vec{v_0} + \vec{a}t[/tex]
    [tex]\vec{r} = \vec{r_0} + \vec{v_0}t+ \vec{a} \frac{t^2}{2}[/tex]

    Now, the trick really is (and that's the essential part) to apply the same procedure in each direction. The procedure i am talking about is projecting each vector along a direction using the triangle-equalities.

    For example : in the x-direction you will have :

    [tex]F_x = ma_x [/tex]
    [tex]v_x = v_{0x} + a_xt[/tex] and
    [tex]x = r_{0x} + v_{0x}t+ a_x \frac{t^2}{2}[/tex]

    [tex]v_{0x} = ||\vec{v_0}||cos( \theta)[/tex]
    [tex]r_{0x} = ||\vec{r_0}||cos( \theta)[/tex]

    the ||.|| denotes the MAGNITUDE of the vector

    You see ? the clue is that each vector can be written as a sum of an x and y component [tex]\vec{A} = A_x \vec{e_x} + A_y \vec{e_y}[/tex]

    [tex]A_x = ||\vec{A}||cos( \theta)[/tex]
    [tex]A_y = ||\vec{A}||sin( \theta)[/tex]

    You will need to be carefull with the signs of the x and y components because those depend of the direction of each component with respect to the actual x and y axis.


    Some exercises :

    1) Derive the equations for both position and velocity when the acceleration is [tex]\vec{a} = 0 \vec{e_x} - g \vec{e_y}[/tex]... Thus a_x = 0 and a_y = -g...This should ring a bell...



    2) So, when a force is given, like : F = m(2e_x - 9.81e_y) and the initial position has components r_0 = 2e_x + 6e_y and the initial velocity v_0 = 6e_y, can you write down the equations for both velocity and position in each direction ???

    In the end , you must realize that this system is very easy because, nomatter how complicated the force may look, the procedure to determin both position and velocity as a function of time is always the same: Projecting the vectors along the given directions. Once, you have done that, you can do almost anything with these formula's...

    regards
    marlon
     
    Last edited: Apr 18, 2005
  2. jcsd
  3. Apr 18, 2005 #2
    Motion on an upward incline

    Once you understand the basic notions of how Newtonian mechanics work, it 'should' be quite straightforeward to apply these rules for various situations. Let us start with the most famous one : motion on an incline. I will post some additional pictures later on but for the beginning it is very important to realize that the incline is upward to the right, the x-axis is along this incline and the y-axis is perpendicular to the incline.

    We will have four forces:

    1) gravity : mg along BOTH x and y axis (realize this !!!)
    2) friction : [tex]{\mu} N[/tex] along the x-axis (in opposite direction!!!)
    3) normal force N along the y-axis
    4) Force F ,along the x-axis,with which we push the object of mass m, upward the incline

    F is pointed upward along the incline so :
    [tex]\vec{F} = F_x \vec{e_x}[/tex]

    The friction force is pointed in the opposite direction of F (you know why ???) so : [tex]-{\mu} N \vec{e_x}[/tex]

    Gravity (and this is the most difficult part) is along both x and y axis
    [tex]m \vec{g} = -mgsin(\theta) \vec{e_x} -mgcos(\theta) \vec{e_y}[/tex]
    Make sure that you know how to derive these formula's when drawing a free body diagram.

    So along the x-axis we have :

    [tex]ma_x = F_x -mgsin(\theta) - {\mu}N [/tex]

    If a_x were zero, then we really are calculating the Force F so that the actual motion will have a constant velocity. Do you get that ???


    Along the y-axis we have :

    [tex] 0 = -mgcos(\theta) + N[/tex]

    Do you realize why F_y = 0 ???

    Now, you can calculate N from the second equation and then plug it into the first one. Now you know everything along the incline (x-axis).

    regards
    marlon
     
    Last edited: Apr 18, 2005
  4. Apr 18, 2005 #3
    Motion on an incline : part 2

    Once you have understood how to work with the motion on an incline, we can elaborate. It is essential to realize that this way of working is completely the same as the projectile motion. What we do is this :

    1) We put the x-axis along the incline and the y axis perpendicular to it. We chose this specific configuration because the formula's will be more easy, however i suggest you do the math when the x-axis is chose horizontally and the y axis vertically!!!

    1) We start to draw the forces at hand

    2) We determin both x and y-components of each force, using the triangle identities

    3) Then we apply Newton's second theory IN EACH DIRECTION. Just add up all components (which can be both positive and negative). The ma on the left hand side denotes the acceleration (up to a constant m ofcourse) that the object will have when all forces of the free body diagram are applied. If the object must remain in the incline, you automatically know that the LHS for the y-direction (ie ma_y) MUST be equal to zero. The reason for this is that the object's position will not change in the y-direction, thus no acceleration.

    4) From 3) you know the acceleration in each direction. This acceleration (ie the a_x and a_y of the left hand side if Newton's second law) can then be substituted in the formula's of the very first post:

    [tex]\vec{v} = \vec{v_0} + \vec{a}t[/tex]
    [tex]\vec{r} = \vec{r_0} + \vec{v_0}t+ \vec{a} \frac{t^2}{2}[/tex]

    When an initial velocity and position have been given, you know everything there is to know about the motion.

    regards
    marlon
     
    Last edited: Apr 18, 2005
  5. Apr 18, 2005 #4
    motion on an incline : part 3

    Now some exercises :

    1) The object's mass is 1 kg
    2) The initial velocity has the following components (7,8)
    This means that v_x = 7 and v_y = 8
    3) The initial position is (0,0) : the origin
    4) The incline is directed upward to the right and the x-axis is along it and the angle with the horizontal is 20°
    5) Given these data, can you calculate how long the object will travel along the incline (assume no friction) ?
    6) How far will it get ?


    I suggest you try solve this in two situations :

    1) when you chose the x-axis along the incline
    2) when you chose the x-axis to be horizontal

    In both cases, you must obtain the exact same results and this proves that the way you chose the reference frame is irrelevant to the actual physical results... Nice isn't it ???

    Another incline problem :
    https://www.physicsforums.com/showthread.php?t=71870&highlight=incline
    https://www.physicsforums.com/showthread.php?t=71354&highlight=incline
    The last one also illustrates how you can crack these problems using conservation of energy...We'll talk about that later

    More to follow later...Then we will conceptualized conservation of energy, work, kinetic and potential energy and we will apply these notions to various situations. I will also add more exercises...

    regards
    marlon

    ps : i will continue tomorrow, gotta go now...if there are remarks, i'd appreciate it if you would PM-me, otherwise this thread will get mixed up with many different kinds of posts... I suggest we keep on following the trajectory that i wrote down in my first post

    thanks
     
    Last edited: Apr 18, 2005
  6. Apr 19, 2005 #5
    Some stuff on energy

    Moving on...we stop at the notion of energy in mechanics. I am very sure that you have heard of both kinetic and potential energy. it is not the intention to introduce those concepts because i am sure that will be done in your textbooks. What can we do with these things ?

    Well if we look back at our incline, i could have asked what the kinetic energy of the object will be, after two seconds of motion. The clue really is that once you followed the procedure sketched in the above posts (ie once you know both the position and the velocity as a function of time) it is really easy to determine the kinetic energy. In our two dimensional example, the velocity has two components v_x and v_y and [tex]v^2 = v_x^2 + v_y^2[/tex]. So all you have to do is fill in the time at which you need to know the kinetic energy.

    Again , the same philosophy arises : we calculate the components of the velocity-vector along the two directions x and y. For each component we can write down the actual expression for velocity as a function of time :

    [tex]v_x = v_{0x} + a_xt[/tex]
    [tex]v_y = v_{0y} + a_yt[/tex]

    Then fill in the given time t and calculate the formula for v²...That's all


    So here is a nice question for you :

    Suppose an object starts moving up an incline which makes an angle of 30° with the horizontal, the object's mass is 10 kg. Assume a friction force of 10 Newton and ofcourse there is also gravity acting on the object. If the initial velocity is 20 m/s along the incline and the initial position is the origin, what will the kinetic energy be after 2 and 5 seconds of motion. The incline is directed upward to the right hand side.

    Hints : be sure that you followed the illustrated procedure that we have sketched in previous posts. Chose an x and y-axis, write down all forces involved and make sure you get their directions with respect to the axis !!! Then calculate all force-components along both directions x and y and substitute these expressions into the equations for position and velocity...

    regards
    marlon
     
  7. Apr 19, 2005 #6
    Work

    First of all, let's remind ourselves what a scalar product is:

    We take two vectors, each with two components (the formula will be the same for any number of components)

    [tex]\vec{A} = A_x \vec{e_x} +A_y \vec{e_y} [/tex]
    [tex]\vec{B} = B_x \vec{e_x} +B_y \vec{e_y} [/tex]

    The scalar product of A and B is defined in TWO ways as :

    [tex]\vec{A} \cdot \vec{B} = A_x * B_x + A_y * B_y[/tex]
    [tex] \vec{A} \cdot \vec{B} = \sqrt{A_x * B_x + A_y * B_y} * cos(\theta)[/tex]
    Where theta is the angle between the two vectors. depending on what has been given for an exercise, you can use one of these formula's...


    We will need the scalar product in order to define the WORK W.

    [tex]W = \int \vec{F} \cdot d \vec{r}[/tex]

    This equation calculates the work done by a force F when the object is displaced over some distance dr. But in most cases F will be constant during the displacement and the formula therefore reduces to [tex]W = \vec{F} \cdot \vec{r}[/tex]

    So basically this means that you can calculate the work in TWO WAYS :
    Suppose that F has components (F_x,F_y) and r has components (x,y)
    1)W = [tex]\vec{F} \cdot \vec{r} = F_x * x + F_y * y[/tex]
    2)W = [tex] \vec{F}\cdot \vec{r} = \sqrt{F_x * x + F_y * y} * cos(\theta)[/tex]


    An example : Suppose the force is [tex]\vec{F} = 2 \vec{e_x} -6 \vec{e_y}[/tex] and the displacement is [tex]\vec{r} = -3 \vec{e_x}[/tex]. then it is quite easy to calculate the work [tex]W = 2 * (-3) + 6 * 0 = -6 J[/tex]

    Did you notice how i used the formula for the scalar product ? Also, the position has no y-component, so that is why i wrote the 0 in the expression for W.

    I realize that this example is very easy, but once again you will recognize our GENERAL STRATEGY : MAKE SURE YOU KNOW THE COMPONENTS OF EACH VECTOR INVOLVED. In this case, we needed the x and y-components of both F and r...

    Here is another exercise : Suppose a 2 kg object is sliding down an incline that makes an angle of 20° with the horizontal. Assume no friction and only gravity is the force that makes the object go down. how much work is done by this force when the object is displaced over a distance of 10 meters ?

    HINT : I suggest you try to solve this problem in two ways, according to the two definitions of the scalar product. The result must ofcourse be the same. You will see that one approach requires less calculations. It is the intention that, in the end, you will get an intuitive feeling that makes you see what approach is most efficient in a particular case where there are multiple ways of solving a problem... Good luck
     
    Last edited: Apr 19, 2005
  8. Apr 19, 2005 #7
    Kinetic energy Theorem

    We ended the discussion on Work by noticing that one of the most important aspects of solving problems is the talent to see which of the different possible approaches requires the least amount of calculations in order to solve any given problem. Now, we shall discuss two such 'systems' that seriously reduce the number of necessary calculations.

    The Kinetic energy Theorem states that : [tex]W = E_k^{END}-E_k^{BEGINNING}[/tex]
    So if you know the kinetic energy at the beginning and at the end of some motion, you automatically know the Work. Then, if you know the Force responsible for this motion, you can directly deduce the distance that the object travelled because of the definition of work. Some examples :

    A 0.3 kg tennis ball is dropped from rest at a height of 3.6 m onto a hard floor.

    a) What is the speed of the ball at the instant of contact with the floor?
    ANSWER sqrt( 2*9.81*3.6) (Make sure you can calculate this for yourself)

    b) A flash photograph shows that the ball is compressed a maximum of 0.6 cm when it strikes the floor. Calculate the force with which the ball is compressed :

    Well, when the ball touches the ground it is compressed 0.006m. At the beginning, the velocity is 8.4m/s (answer of a)). This is the moment when the ball first touches the ground. At the end, the ball is compressed and the kinetic energy has become 0 (because the ball is actually on the ground, it does not move !!!).

    So the difference in kinetic energy is just the kinetic energy of the ball when it touched the ground. This is easy to calculate. This E_k is equal to the work done by this force : W = d*F = E_k. Since d = 0.006, you can solve this equation for F and that's your answer....



    marlon
     
    Last edited: Apr 19, 2005
  9. Apr 19, 2005 #8
    Conservation of total energy

    The conservation of total energy, ie the sum of kinetic and potential energy, is very well known. This law applies to the conservative systems, which means that a conservative force is responsible for the actual motion from a point A to a point B. The definition of 'conservative' goes like this : the work done by a conservative force does not depend on the path followed by the object in order to go from A to B. Total energy is also referred to as mechanical energy. examples of conservative forces are gravity, the spring force,...

    This law enables us to seriously reduce the number of required calculations. the most important aspect is that you know how to work with a conservation law. The clue is that you can calculate the sum of both kinetic and potential energy in TWO different situations. These sums will be equal to each other thanks to the actual conservation law. An example :

    Suppose we have an object that moves down an incline which makes an angle of 60° with the horizontal. Initially the object is at rest. The only force that acts on the object is gravity. This is a crucial aspect because now you know that we can actually apply this law because gravity is a conservative force. So, in exercises, you will always need to check whether you can apply conservation of energy by checking if the forces at hand are indeed conservative. The object's initial position is on the y-axis at height h. What will be the velocity when the object has moved off the incline (ie when it is on the horizontal) ?

    Well, the strategy is always the same. First we look at the beginning and we calculate both kinetic and potential energy :[tex]\frac{mv^2 = 0}{2} +mgh[/tex]. You know that the potential energy associated with gravity is mgh, where h is the vertical height. The kinetic energy is zero because v=0 (object is initially at rest).

    Then we look at the end. the object is now on the horizontal x-axis. hence, the potential energy is 0 (h = 0) and the kinetic energy is just [tex]\frac{mv^2}{2}[/tex] where v is the velocity we are looking for.

    Thanks to the conservation of total energy both sums are equal : [tex]mgh = \frac{mv^2}{2}[/tex]. Now, solve this equation for v and there is your answer...


    Here's an exercise : You know that the potential energy associated with a spring that has been compressed over a distance x is equal to [tex]\frac{kx^2}{2}[/tex], where k is the spring constant. Now, a spring gun is compressed over a distance of 3,2 cm and a ball of mass 12g is inside the gun. With what speed will the ball leave the barrel when the gun is fired ? Assume k to be 7.5 N/cm and assume no friction inside the barrel. Your answer should be v = 8 m/s

    Another one :
    A 61 kg bungee cord jumper is on a bridge 45 m above a river. In its relaxed state, the elastic bungee cord has length L = 25m. assume that the cord obeys Hooke's law with a spring constant of 160N/m. If the jumper stops before reaching the water, what is the height of her FEET above the water at her lowest point. your answer should be 2.1m. What is the net force on her at her lowest point (in particular is it zero?)


    regards
    marlon
     
    Last edited: Apr 19, 2005
  10. Apr 19, 2005 #9
    Important remark on energy conservation

    When working with energy conservation, you should ask yourself four questions :

    1) For what system is total energy conserved ?
    Well, for conservative systems. How can you recognize those ? Well, you should be able to separate the system from its environment. You should be able to draw a closed surface such that whatever is inside the surface is your system and whatever is outside is the environment. In the last exercise, the system is the jumper + cord and the environment is the earth.

    2) Is friction or drag present ?
    If so, total energy will not be conserved because these are non-conservative forces. We will see an adapted formula for these cases : [tex]\Delta E_k + \Delta E_p = \Delta E[/tex] where the RHS is the work done by the non-conservative forces. In case of a friction force, this will be [tex]\Delta E = -F_{friction}d[/tex]

    3) Is your system isolated ?
    Conservation of total energy only applies to isolated systems. This means that no external forces (forces exerted by objects outside the system) should do the work on the objects inside the system. For example, in the first exercise, if you chose the ball to be your system,it will not be isolated because the external spring force works on it. Thus total energy of the ball will not be conserved. the system must be the gun AND the ball. the environment is the earth.

    4) What are the initial and final states of your system ?
    You already know why you need those... :smile:

    regards
    marlon
     
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