1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Various forces on various objects

  1. Nov 24, 2016 #1
    1. The problem statement, all variables and given/known data
    sorry, me again.
    The following figures show various forces acting on gliders of various masses at rest on a frictionless track. In each case, the force is applied for 30cm. The choice listing the examples in decreasing order for the increase in kinetic energy is {-4J < -2J < 1J}
    7f707f1f350817579ac49d29ff86bcfc.png

    answer:(a) C > A=B > D (b) B=C > D > A (c) C > A > B > D (d) B > C > D > A (e) B > C > A > D

    2. Relevant equations
    W=FD
    F= MA

    3. The attempt at a solution
    I dont clearly understand the questions here, is it asking for sorting it from most work to least? (cant ask teacher since he nevers check his email). if so wouldn't it be D>c>b>a since the mass of it all is combine on d, so it requires more force? I dont understand how in the answer, a block can be = to another block, of the other block (combined) mass is more?
     
  2. jcsd
  3. Nov 24, 2016 #2

    Doc Al

    User Avatar

    Staff: Mentor

    What determines the change in kinetic energy?
     
  4. Nov 24, 2016 #3

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Yes. It's asking you to put them in order of the KE gained. So which one gains the most KE? By conservation of energy the KE gained = work done by the applied force.

    No. Treat all of the drawings as separate experiments. They don't collide.

    One can equal another if it gains the same KE as another.
     
  5. Nov 24, 2016 #4
    mass and velocity, but how can I find velocity without the time? I'm looking at all the velocity equations I know and it all depend on time.
     
  6. Nov 24, 2016 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Forget all that. Think of the work-energy theorem.
     
  7. Nov 24, 2016 #6

    Doc Al

    User Avatar

    Staff: Mentor

    When I say "forget all that" I am being a bit facetious. You could do it that way, but that's way harder than necessary. You are given the force applied and the distance -- so use it.
     
  8. Nov 24, 2016 #7
    Ok, so I calculated out the work done on each block seperately, A = 500J B=2000J C=1000J D= 1125J, so it's B>D>C>A, but that's not a choice, the closest is B=c>d>a and the reason why b=c is because c is half the weight as b, and if it was the same it would ACTUALLY be equal? am I correct?

    edit: wait... why am I multiply force by the mass????????? ignore this then....
     
  9. Nov 24, 2016 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Hint: The distance is the same for each, so just arrange by force.
     
  10. Nov 24, 2016 #9
    oh wow... that's so much easier
     
  11. Nov 24, 2016 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    As Doc Al writes, you do not need to find the velocity. But if you insist on taking that route....
    There are five standard variables in the equations for constant acceleration, the symbols for them spelling the acronym SUVAT.
    Any four can be related by an equation, so there are five standard equations, each omitting one variable. In particular, one omits t. That one is effectively an energy equation, but with mass cancelled out.
     
  12. Nov 24, 2016 #11
    what if say that it's being pushed for 1s? can we ignore 1s too since it's all being pushed for 1 s?
     
  13. Nov 24, 2016 #12

    Doc Al

    User Avatar

    Staff: Mentor

    Nope. Since each will have a different acceleration, the distances will be different even though the time is the same. But in that case, you might want to use the impulse momentum theorem to help find the change in energy. (It will take a few steps.)
     
  14. Nov 24, 2016 #13
    So I would have to find accel from the force, from there find the velocity of 1s, then from there find the impulse and sort the impulse in increasing order?
    if so my increasing order should be B=C>D>A does that sound right?
     
    Last edited: Nov 24, 2016
  15. Nov 24, 2016 #14

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Correct.
    The easy way is to use KE=work=force*displacement.
    The hard way would be to use Newton's law and the equations of motion (SUVAT) to calculate the final velocity for each. Then use KE=0.5mv2
     
  16. Nov 24, 2016 #15
    if I were to use force*displaycement, what would happen to the time then? where would that go?
     
  17. Nov 24, 2016 #16
    I'm confuse, still on the same model as original post, it's ask for " increasing order for the increase in momentum is" but it's giving me
    (a) C < B < D < A (b) C < A=B < D (c) B=C < D < A (d) C < B=D < A (e) A < C < D < B and the closest to my answer is C, but the signs are wrong... I think.
    Are the sign correct?
     
  18. Nov 24, 2016 #17

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    One of those answers is correct, but it is not C.

    I've put the thread back to Unsolved.
     
  19. Nov 24, 2016 #18
    it's not? I thought it was list the impulse in increasing order
     
  20. Nov 24, 2016 #19

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Sure, but you first have to calculate the impulses. I see no working by you in that regard.
    I did not understand this part of an earlier post:
    Where do you get 1s from? In each case, the force is applied over the same distance. That distance will be covered in different times. So you cannot multiply the force by 1s to get the impulse.
     
  21. Nov 24, 2016 #20
    from
    and I did the work on paper. I didnt multiply it by the seconds, I first found the accel by diving the force by mass, then from there use A=Δv/Δt for the interval of .1s to get velocity final, and from there I did mass time Δv (v final since initial is 0) which gave me the [change in momentum]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted