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Homework Help: Simple physics question on conservation of energy

  1. Nov 9, 2011 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    total initial energy = total final energy

    3. The attempt at a solution
    what i got
    (i) 33.55J
    (ii) 98.1J
    (iii) 27.66J
    I am unable to obtain (iv). having block B in mid air and i have no height given to me, I cannot input any height for my initial "potential energy = mgh". So i assumed height to be 0 for initial block B and got a height of 0.34m for block A (h= sin 20)...

    after trying working out all the values, i got a negative kinetic energy? which is totally wrong!!!

    P.S: there is no answer sheet, so i would not know if all my previous answers are correct as well.
  2. jcsd
  3. Nov 9, 2011 #2
    You can do part 4 in at least two ways. First, you could find the net force on the system by summing all the forces acting on block A. Then use Newton's law to determine acceleration followed by the relation that equates velocity to acceleration and distance. Solving for velocity^2 will then allow you to determine the kinetic energy.

    Or you could utilize energy conservation which says the sum of the initial potential and kinetic energies equals the sum of the final kinetic and potential energies plus the work done by the system.

    You'll get the same answer each way. And when you do, you'll know you are correct.
  4. Nov 9, 2011 #3
    i will try out the 1st way using newton's law. as for using energy conservation, i have tried out and have obtained a negative K.E.? I think the reason was because i had no initial potential energy for block B and no final potential energy for block A. ?
  5. Nov 10, 2011 #4
    I'll give you a hand with the energy solution when you get around to it.
  6. Nov 10, 2011 #5
    attempt using newton's law:
    F = ma = 200-27.66+33.55-98.1 (these 3 values are taken from the force that i had calculated in the previous parts (i)(ii)(iii))
    20a = 107.79
    a = 5.3895m/s^2

    v^2 = u^2 + 2as
    v = 3.28m/s

    K.E. = 1/2 mv^2
    K.E. = (0.5)(10+10)(3.28)^2
    = 107.79J

    is that correct?
  7. Nov 10, 2011 #6
    That is correct. Now see if you can do it using energy considerations.
  8. Nov 10, 2011 #7
    i think the problem with this question for me would be getting the initial potential energy.
    here is my working:
    total initial energy = total final energy
    1/2mv^2 + mgh = 1/2 mv^2 + mgh

    since system is at rest initially, 1/2mv^2 = 0
    total initial potential energy = mgh(a) + mgh (b) = (10)(9.81)(0.34) + (10)(9.81)(0)
    = 33.354N

    total final energy = K.E. + mgh (since we are looking for total kinetic energy, i left the K.E alone, only putting in the values for mgh.)
    K.E. + mgh(a) + mgh(b) = K.E + (10)(9.81)(0) +(10)(9.81)(1)
    = K.E + 98.1

    initial = final
    33.354N = K.E + 98.1
    K.E. = -64.746N (negative value?)

    if it wasn't for the negative value which i got, i would have thought that i was on the right track.
    Do tell me if i am wrong, but i cannot assume that my initial height for block B = 0?
  9. Nov 10, 2011 #8
    In looking at your solution I see that you have omitted the work terms. Look at the following equation.

    KE1 + PE1 = KE2 + PE2 + Work done by/on system


    KE2 = KE1 + PE1 - PE2 - Work done by/on system

    KE2 = KE1 + delta(PE) - Work done by/on system

    See if this helps. Hint: Work done by/on system has two components.
  10. Nov 10, 2011 #9
    PS: There is work done BY the system and work done ON the system. They carry different signs.
  11. Nov 10, 2011 #10
    KE1 + PE1 = KE2 + PE2 + Work done by/on system
    0 + 33.354 = KE2 + 98.1 (from previous working)

    work done by/on system:
    since the force is towards the right:
    + given force + block A - Block B (string tension) - friction
    = 200 + 33.55 - 98.1 - 27.66

    KE1 + PE1 = KE2 + PE2 + Work done by/on system
    0 + 33.354 = KE2 + 98.1+ 200 + 33.55 - 98.1 - 27.66
    33.354 = KE2 + 205.89
    KE2 = -172.536N (still a negative?)

    i am sorry to have bothered u so much but i still can't get the answer.
    also, KE2 = KE1 + delta(PE) - Work done by/on system --> what u mean by delta (PE)?
  12. Nov 10, 2011 #11
    OK, I'll show you my work.

    KE2 = KE1 + delta(PE) - Work done by/on system, delta(PE)=PE1-PE2

    KE1 = 0 because no motion initially
    PE1-PE2 = 0 - 10(9.81)(1) + 10(9.81)(sin(20)) = -64.54 J
    Work done by system = 10(9.81)(0.3)(cos(20)) = 27.7 J Work done by system
    Work done on system = -200(1) = -200 J Work done on system

    KE2 = PE1-PE2 - Work
    KE2 = -64.5 - 27.7 -(-200) = 107.8 J
  13. Nov 10, 2011 #12
    I get it now. just to clear some doubts,
    work done BY system = positive
    work done ON system = negative
    - the signs are fixed irregardless of direction? for example, if i were to exert a second force of 100N on block B downwards or upwards, since it is work done ON the block, it will be a negative?)

    work done by gravity are not included? why?
    Last edited: Nov 10, 2011
  14. Nov 10, 2011 #13
    Work done on the system is negative. Work done by the system is positive and it does not matter which direction it's in. Work done on the system is negative and it does not matter which direction it's in. If you always go back to the first law of thermodynamics, you won't go wrong. But I don't know if you've suffered through thermo yet.

    The work done by gravity is accounted for by the change in PE.

    Another way to look at is that you are doing an energy balance on the system, before and after some sort of happening. The LHS represents before; the RHS represents after the happening. If the system has done work, its total energy (RHS) should be less. If work is done on the system its total energy (RHS) should be more so the double minus signs make the term positive.
  15. Nov 10, 2011 #14
    ok... thanks alot.
    FYI: i am an electrical student, and i have no idea why am i learning this. struggling in my circuit though. =(
  16. Nov 10, 2011 #15
    When I was a mechanical engineering student, we had to take a year of EE courses. We called it EE for non-EE's. Professors had fun with us because we were never sure about what we were doing in those two classes.
  17. Nov 10, 2011 #16
    wait wait...
    i got confused again.

    the equation is:
    KE1 + PE1 + W = KE2 + PE2
    KE1 + PE1 = KE2 + PE2+ W
  18. Nov 10, 2011 #17
    KE1 + PE1 = KE2 + PE2 + W

    where work done by the system is positive and work done on the system is negative. So if you are solving for KE2 (KE1 = 0), you get this:

    KE2 = (PE1 - PE2) - W

    Now, if the system is doing the work W > 0 so KE2 is less. Doesn't that make sense? If you are doing work, you'll have less KE assuming delta PE is constant. You don't ever get something for nothing.
    If work is being done on the system, W < 0, so KE2 is larger.
  19. Nov 10, 2011 #18
  20. Nov 10, 2011 #19
    The piece in wiki doesn't say anything outright about the sign of the work term (what's >0 and what's < 0. If one reverses the sign convention I am using, then you can put the W on the LHS. I always go back to the First Law of Thermo which is the way I learned it. If you feel more comfortable with W on the LHS, then work done by the system would be negative and work done on the system would be positive.

    Among the last few statements in wiki are:
    W = KE2 "where W is the work done on the body under consideration"

    F d = 1/2 m v2

    This implies that work done on the system is positive because KE is always positive. I use the opposite sign convention for work. Use whichever makes you more comfortable. Just don't get it all mixed up when it's time for a test.
  21. Nov 10, 2011 #20
    ok. thank you.
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