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Homework Help: Work in a circular track question

  1. Apr 16, 2008 #1
    1. The problem statement, all variables and given/known data

    This question came up recently, and i personally have been weak regarding work and energy, i have only just started to understand newtons laws sufficiently. There is a scan of the question attached. Could i please have an explanation this question?

    A 0.5-kg mass is projected down a rough circular track (radius = 3.1 m) as shown. The speed of the mass at point is 3.1 m/s, and at point , it is 6.3 m/s. How much work is done on the mass between and by the force of friction?

    2. Relevant equations

    i tried to use the work energy theorem ie W = ΔK but it didn't feel correct

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Apr 16, 2008 #2


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    The work energy theorem you have noted applies to the total or net work done by all forces acting on the object, including gravity. You just want the work done by the friction force. Can you identify another equation that will identify the work done by friction alone?
  4. Apr 17, 2008 #3
    it wouldn't have anything to do with gravitational potential energy would it?

    im not too sure however
  5. Apr 17, 2008 #4
    Matter of fact, it has everything to do with GPE.

    Since no initial velocity is given to it i.e. the object is pushed gently over the track.. it has no initial kinetic energy.

    Firstly, take the potential energy at the height of 'A' to be zero. Then, calculate the GPE at the starting point. Once you've done that.. use the velocity at point 'A' to calculate the final Kinetic Energy at that point. This Kinetic energy should be equal to the GPE at the starting point. If not, then it should be less and the difference between the KE and GPE is the work done by frictional force.

    Use the same principal to calculate the work done along the path AB.
  6. Apr 17, 2008 #5
    hey guys, i worked it out and got close to -30J. It seems a bit better now.

    Is this correct?
  7. Apr 17, 2008 #6
  8. Apr 17, 2008 #7
    hi rohanprabhu,
    which equations did you use in ur calculation?
  9. Apr 17, 2008 #8


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    Hi rohanprabhu,

    The expression you typed into google was

    0.5 * ((3 * 9.8 * 3.1) - (0.5 * ((3.1^2) + (6.3^2)))) = 33.24500

    That doesn't look right to me. The work done by friction, since it is the only non-conservative force, will be equal to the change in the total energy, so

    W = E_f - E_i
    or another form is
    W = (KE_f - KE_i) + (PE_f - PE_i)

    It looks to me like there are several errors in your expression.
    Last edited: Apr 17, 2008
  10. Apr 17, 2008 #9


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    Hi pcwizz,

    I got a different answer. What numbers were you using?
  11. Apr 17, 2008 #10
    @alphysicist: Well.. I get your point. And i don't know why.. but i computed the work done by friction first till the point 'A' and then till the point 'B' and then added them. I don't know why I did it in 2 parts.. but here's what i did:

    Let the GPE at the height of 'A' be 0. Then, at the starting point,

    E = mgr

    [m=mass of object; r=radius]

    at the point 'A', let the work done by friction be [itex]W_{s1}[/itex]. Then,

    \frac{1}{2} mv_{a}^2 + W_{s1} = mgr


    W_{s1} = mgr - \frac{1}{2} mv_{a}^2

    At the point, B the GPE = -mgr. Let the work done by friction till here be [itex]W_{s2}[/itex].

    \frac{1}{2} mv_{b}^2 - mgr + W_{s2} = \frac{1}{2} mv_{a}^2


    W_{s2} = mgr + \frac{m}{2}(v_{a}^2 - v_{b}^2)

    hence, we get:

    EDIT: I made an arithmetic mistake in the next step.. sorry for that
    and thanks alphysicist for pointing it out

    W_s = W_{s1} + W_{s2} = 2mgr - \frac{mv_{b}^2}{2}

    which when i substituted the values.. i got the answer as 20.46J. Sorry for that again...

    @OP: the method which i used is not a good method for solving this problem. It takes more time and does not have any advantage at all. To solve this, use the GPE and KE at only the initial and final points. I should've understood that there is an error when i saw that the final answer I got earlier was dependent on the velocity at a.
    Last edited: Apr 17, 2008
  12. Apr 17, 2008 #11


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    I think you might have misread the problem. It looks like you are assuming that we know the object has speed=0 at the highest point of the circular part of the track, and then you followed a procedure that would give the work done by friction between the highest point of the track and the lowest point.

    (However, although friction does not always need to do negative work, it is doing negative work here. So your result needs to be -20.46J. For example, I think your expression for the work [itex]W_{s1}[/itex] done between the top and point A should be

    W_{s1}=\frac{1}{2} mv_a^2 -mgr
    if you assume v=0 at the top and set the height h=0 at point A. The work is related to the change in energy, which is final energy - initial energy.)

    But actually in this problem they only want the work done between points A and B, so they only want what you are calling [itex]W_{s2}[/itex]. But I think it needs to be the negative of your expression because:

    [tex]E_a= m g h_a + \frac{1}{2} m v_a^2[/tex]

    [tex]E_b= m g h_b + \frac{1}{2} m v_b^2[/tex]

    and so

    [tex] W_{a \to b} = E_b - E_a[/tex]

    Plugging in the values and choosing [itex]h_a=0[/itex] would give

    [tex] W_{a \to b} = (m g h_b + \frac{1}{2} m v_b^2) - (\frac{1}{2} m v_a^2)[/tex]
  13. Apr 17, 2008 #12
    Whether I get the work done by friction as negative or positive depends upon my reference. In my method of solving it, I have taken that the work done by the frictional forces and the increase in KE should add up to the GPE. What you have done is that the difference in the GPE and KE should be accounted for by the work done by the frictional force. Both ways are correct with the difference that you consider that the work done on the system to be negative [which is perfectly what they do in the thermodynamics convention] and I took that to be positive.

    And yes, I have taken the initial velocity as 0. I didn't notice any part where they asked the work done between point 'A' and 'B' and I thought it was for the entire track and hence assumed so.
  14. Apr 17, 2008 #13


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    In the link to the picture the problem stated to find the work done between A and B. In the original post text that was pasted to the forum the symbols for A and B did not show up.

    I don't believe you have the freedom to change the sign of the work. The work done on the mass is always positive if it's energy is increasing, and is negative is it's energy is decreasing. The mass's energy is decreasing, so the work done on it must be negative.

    You can also see this by the definition of work [itex]W=\vec F\cdot \vec d[/itex]. As the mass slides down the frictional force is always opposite the displacement direction and so the work is negative.

    Instead of the work done by friction, were you perhaps calculating the increase in thermal energy? They are negatives of each other, but it's important to keep them separate. (If a problem asked for the work done by gravity, we would not be able to give the change in gravitational potential energy even though they are negatives of each other.)
  15. Apr 18, 2008 #14
    hi alphysicist and rohanprabhu,
    does h and r stand for radius? because i plugged in my data into both of your equations and did not get a right answer.
    my data is a bit different (m=1.1, r=3.2m, V1=3.6m/s, v2=8m/s) but the question is the same. i got around -50J to -60J using both your equations
    however the answer is a lot smaller than this!
    can you help me?
  16. Apr 18, 2008 #15
    I get a much smaller number with alphysicist's last equation (after realizing that [itex]h_b[/itex] was negative.
  17. Apr 18, 2008 #16


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    Hi fizwiz,

    The post by kamerling will solve your discrepancy I believe.

    To explain what happened, my full formula would be

    [tex]W_{a\to b} = E_b - E_a [/tex]

    [tex] W_{a \to b} = \left( \frac{1}{2} m v_b^2 + m g h_b\right) - \left( \frac{1}{2} m v_a^2 + m g h_a\right)[/tex]

    In rohanprabhu's first post he wanted to choose [itex]h_a =0[/itex]. If you choose that, then, since point B is below A, you would need to use [itex]h_b=-3.2[/itex] (if 3.2 meters was the radius).

    If you had chosen point B to be at height zero, so that [itex]h_b=0[/itex], you would get the same answer; the only difference would be that since A is above B the height would be positive, so in that case [itex]h_a=+3.2[/itex].

    Usually when I work these problems, I choose the lower point of the two I'm using for the energy equation as having height=0; that way you won't have a negative sign for the height of the other point and it's one less source of error. It's just that in this thread since the choice [itex]h_a =0[/itex] had already been made I did not want to confuse things.
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