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Work and Energy of falling mass attached to mass with friction on incline.

  1. Oct 8, 2010 #1
    1. The problem statement, all variables and given/known data
    [PLAIN]http://rawrspace.com/physics.GIF [Broken]


    2. Relevant equations
    v=v0+at
    x=x0+v0t+(1/2)at2
    W=[tex]\Delta[/tex]K=K2-K1

    3. The attempt at a solution
    For the first part of this problem I attempted it this way which I think is completly wrong.

    I rearranged the first two equations to get 2a(x-x0)=(v2-v02)

    Then i multiplied by (m+M) on both sides and ended up with

    (m+M)(x-x0)=(1/2)(m+M)(v2-v02)

    Which I don't think is what they are looking for. I believe it is because I am using the wrong equations. How would I approach this with the other equations. I know that Gravitational potential energy is U=mg[tex]\Delta[/tex]y and kinetic energy is (1/2)mv2

    I also know that if a free body diagram were drawn , you would have the tension set equal for them (however it doesn't start at rest so this might not be true, correct ?) that would give you mg[tex]\Delta[/tex]y = (1/2)mv2 + Wf

    where Wf is the work done by friction.

    I know total work in the system is Wg = Wa - Wf

    I just feel like this has confused me quite a bit and I am not sure which route to approach it from. If someone could shed some light on how to do this problem and how to determine the work done by gravity and friction, I would appreciate it.
     
    Last edited by a moderator: May 5, 2017
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  3. Oct 8, 2010 #2

    vela

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    You're complicating the first part way too much. Since the two blocks are connected by a string, they have to move with the same speed. Say they initially move with speed vi. What's the kinetic energy of the block with mass m? What's the kinetic energy of the block with mass M? What's their total kinetic energy Ki? Do the same thing with their final speed to find Kf. Subtract the two results to find the change in kinetic energy of the system.
     
  4. Oct 15, 2010 #3
    Alright so for the first part

    k1= (1/2)mvi2 + (1/2)Mvi2
    k2= (1/2)mvf2 + (1/2)Mvf2
    [tex]\Delta[/tex]k = k2-k1
    so
    [tex]\Delta[/tex]k = (1/2)(m+M)[vf2-vf2]

    That satisfies the first question.

    Then it asks for the work done by gravity.
    So we know that there is potential energy for the smaller block which is mgh , it says to assume that the system does not start at rest, so setting up a mec energy equation would be
    energy of m = energy of M
    U1 + K1 = U2 + K2
    Since m starts at a height of h then U1 = mgh
    I believe that since it is not starting from rest then it also has a kinetic energy, so K1 = (1/2)mvi2
    for M , it does not start at the bottom but there is no way to get height that I see, so would U2 = 0 and K2= (1/2)Mvi2

    However , I am thinking that the equation actually results to something like U1 = K2 , or would that only be if it started from rest.
    using each way this is what I get, using the first idea I get

    mgh + (1/2)mvi2 = (1/2)Mvi2
    so mgh=(1/2)(M-m)i2-f

    Normally Wg is given by -mgh ... so would the work done by gravity be
    Wg=-mgh=-(1/2)(M-m)i2+f
    this seems to say that the falling m is equal to the kinetic energy of M and then the initial kinetic energy (while positive in the problem) reduces the work. This would give a negative value for work done by gravity, but it is falling so it is losing energy correct, so a negative value seems correct.

    If you used the second idea then you get
    mgh=(1/2)Mvi2-f
    which means the work would simply be the negative kinetic energy of M , that seems that it would work because the kinetic energy is being generated by the falling mass.
    Again you have a negative value for the work , but that makes since since it falling.

    I believe the first method seems more accurate since we do not know the kinetic energy of the block m.

    as for the last part, it asks for the work done by friction. I know that there is a coefficient of friction between M and the slope, but how do I introduce that into the work equation ?
    Would it be
    mgh=(1/2)(M-m)i2-f
    then just swap mgh and f to solve for the work ?

    thanks for taking a look at my work and letting me know.
     
  5. Oct 16, 2010 #4
    Is anyone able to verify any of the above ?
     
  6. Oct 16, 2010 #5

    vela

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    What exactly do you mean by K1, K2, U1, and U2? Are they the initial and final energies of the system (as you used k1 and k2 in the first part) or are they the energies of mass 1 and mass 2 (as you seem to be using them now)?
     
  7. Oct 16, 2010 #6
    What I mean in the second part is that

    U1 + K1 , the initial gravitational and kinetic energy is equal to U2 + K2 the final graviational and kinetic energy.

    So since m starts at mgh and the system has an initial kinetic energy of (1/2)Mv^2

    that is equal to the final which is Mghsin[tex]\Theta[/tex] ( i believe since the mass M moves up the incline that much) and the kinetic energy final would be (1/2)mv^2

    does that seem wrong, if so , any nudges that would get me to the right formula.... I think that is wrong because I do not have friction involved...it should be I would think....

    Thanks.
     
  8. Oct 16, 2010 #7

    vela

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    You really need to nail down the basics of energy before you try to use energy concepts to solve problems.

    This problem is asking you to calculate the work done by gravity and the work done by friction. So looking up the definition of work would be a good starting point.
     
  9. Oct 17, 2010 #8
    Well work is force times distance cos theta
    For the Work done by gravity it is mgh so would I set

    mgh = Mg[hsin(theta)]-f

    f = Mgcos(theta)

    so
    mh=M[hsin(theta)-cos(theta)]

    is that a step in the right direction ?
     
  10. Oct 17, 2010 #9

    vela

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    You're fine up to here. What do you get when you apply this definition to each block (separately)?
    None of this makes sense.
     
  11. Oct 17, 2010 #10
    well for block m , it would be cos 90 which would be 0 , but I was under the impression that you had to shift the axis so that there is actually a value calculated because if you have it as 0 no work is done. Which is true I suppose since it is moving up and down only.

    As for block M , you would have Mgcos(theta) ,don't I have to multiply this by the distance, which would be h , is that right ? Then you would need to subtract the force of friction wouldn't you ? so -f , the force of friction is mu times the normal force. So -Mgcos(theta) when put into the -f equation is -mu*Mgcos(theta)
     
  12. Oct 17, 2010 #11

    vela

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    What is the definition of theta?
     
  13. Oct 17, 2010 #12
    theta is the angle of the incline.

    I do not see a way to express this in other terms because the only variable of length I have is h .
     
  14. Oct 17, 2010 #13

    vela

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    Sorry, I meant what's the theta in the formula for work? It's not the angle of the incline, even though they're using the same symbol.
     
  15. Oct 17, 2010 #14
    It is the angle between the force and the displacement vector.

    the force is Mg since it is the mass times gravity
    the displacement would be found by calculating how much the block moves because of the force of block m falling. If block m falling travels a distance h, then block M moves a distance h because they are connected by the same string. Since it is on an incline however wouldn't you calculate the displacement for x as hcos([tex]\Theta[/tex]) and y as hsin([tex]\Theta[/tex]) where [tex]\Theta[/tex] is equal to the incline of the ramp.

    That would make the angle between the force and displacement ...

    this might be wrong but looking at the problem here, the force it pulling at the angle of the incline, and the displacement is moving at the same angle, so that would make the angle between the two 0 degrees, which would be 1.

    but that would get me stuck again, because if I calculated the force as being Mg it wouldn't work. However the tension is providing the force pulling it up the ramp. So if I checked the force for block m , it has the tension being equal to -mg , then if I draw a force diagram for block M , it has it's weight pulling down Mg and it has the normal force , -Mgcos[tex]\Theta[/tex] (where this is the angle of the incline) and the force of tension which will be equal to the tension of the falling block , so -mg . Since the only things affecting motion in the x direction are tension and the normal force, they would be opposite which would get mg=Mgsin(theta) since I want the X component of gravity (parallel to the plane)

    Can you let me know if I am way off base ? I am going to tackle this problem again in the morning. I appreciate all the help you are providing, I think this problem is most likely not that hard, something is just not clicking for me.
     
  16. Oct 17, 2010 #15

    vela

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    I'm guessing you're getting confused because you've seen different ways to calculate the work, and you're mixing them up. This is why it's important to understand what the equations mean rather than just blindly plugging stuff in. At the very least, you need to know what the variables in an equation mean.

    The work done by a constant force F acting over a displacement d is

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

    In this problem, for example, if you want to calculate the work done by gravity on the block on the incline, F would be the force of gravity, which has a magnitude of Mg and points straight down, and d would be the vector parallel to the incline, pointing in the direction the block moves and having a magnitude of h.

    Method 1

    You've probably seen two different ways to calculate the dot product, and you can calculate the work by either method. If you know the magnitudes of the vectors and the angle between them, you can find the work using

    [tex]W = \vec{F}\cdot\vec{d} = |\vec{F}| |\vec{d}| \cos\phi[/itex]

    where ϕ is the angle between the two vectors when placed tail-to-tail. For the block on the incline, the angle between the two vectors is ϕ=θ+90, so the work is

    [tex]W = |\vec{F}| |\vec{d}| \cos\phi = Mg \times h \times \cos(\theta+90) = -Mgh\sin\theta[/tex]

    Method 2

    The other way is convenient when you can easily break the vectors up into components. In this case, you'd have

    [tex]W = \vec{F}\cdot\vec{d} = F_x d_x + F_y d_y[/tex]

    where F = Fxi+Fyj and d = dxi+dyj. For the incline on the block, the force of gravity points straight down, so F=-Mgj. As you already noted earlier, the displacement in the x-direction is -h cos θ (it moves in the -x direction) and the displacement in the y-direction is h sin θ, so you'd have d=(-h cos θ)i+(h sin θ)j. You'd then calculate the work as follows

    [tex]W = F_x d_x + F_y d_y = 0\times(-h \cos\theta) + (-Mg)\times(h \sin \theta) = -Mgh\sin\theta[/tex]

    You get the same answer as before, as you should. You should note that the only term that contributes to the answer is the y term. Only the vertical displacement matters because the force is vertical. More generally, you can calculate the work by multiplying the magnitude of the force by the displacement in the direction of the force. In your book or notes, you've probably seen this idea expressed as

    [tex]W = F d \cos \phi = F (d \cos \phi) = F d_\parallel[/tex]

    Method 3

    In inclined plane problems, sometimes you want to use rotated axes, where the x'-axis runs parallel to the incline and the y'-axis is perpendicular to the incline. Let's assume the +x'-direction points down the incline and the +y'-direction points in the same direction as the normal force. In that case, you'd have F=(Mg sin θ)i+(-Mg cos θ)j and d = -hi, so the work would be

    [tex]W = F_{x'} d_{x'} + F_{y'} d_{y'} = (Mg \sin\theta)\times(-h) + (-Mg\cos\theta)\times 0 = -Mgh\sin\theta[/tex]

    This time, only the x-term contributes to the work. Only the component of the force that points along the displacement matters. So you can calculate the work by multiplying the distance the block moves by the component of the force along the displacement. You might have seen this idea written as

    [tex]W = F d \cos \phi = (F \cos \phi)d = F_\parallel d[/tex]

    So you see, you have several different ways to solve the problem. You just have to keep the details straight.
     
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