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Are my answers correct? (newton's laws etc)

  1. Dec 17, 2011 #1
    I have my first physics exam coming up in January. To study I've been doing the past exams from the previous years, but there were are a few questions that I wasn't 100% sure about. I'd be greatful if anyone could tell me if my answers are correct or not :smile:


    1) Which of the following forms of Newton’s Second Law may be used when considering a body of constant mass?

    i) [itex]F=m\vec{a}[/itex]
    ii) [itex]F=m\frac{d\vec{v}}{dt}[/itex]
    iii) [itex]F=\frac{d\vec{p}}{dt}[/itex]

    A. (i)
    B. (ii)
    C. (iii)
    D. (i), (ii) and (iii)
    E. None apply​


    Now, I know that if the mass is constant then equation iii) can be used. That makes me think the answer is C. But then again, if the mass is constant I could technically use i) and ii) if I wanted to aswell? So the answer is D.?



    2) A woman pulls a car 50 m along the ground using a tow-rope attached to
    the front of the car. If she does 1000 J of work, exerting a force of 40 N, at
    what angle is the rope to the ground?
    A. 30°.
    B. 45°.
    C. 60°.
    D. 75°.
    E. 90°.​


    Since she's doing half the work she 'should' be doing, the y-component must be taking half the force, or 20N. So is this really as simple as sinθ = 20/40, meaning the answer is A. 30°?



    3) The potential energy for a two-dimensional force is given by the formula
    U= 6x2 y + 6y4 . What is the force acting at the point (x, y)?

    A. F = (12x)i + (6y3 )j N.
    B. F = (6xy + 6y4 )i + (6x2 + 18y3 )j N.
    C. F = 6x3 y2 + 6xy5 N.
    D. F = 12xy + 24y3 N.
    E. F = (-12xy)i + (-6x2 - 24y3 )j N.​

    Differentiating with respect to X gives 12xy, and with respect to Y gives 6x2 + 24y3. The only answer with these values is answer E, but E gives negative forces. Is this because gravitational potential energy (U) is negative? Or is my answer just wrong?
     
  2. jcsd
  3. Dec 17, 2011 #2
    for question 1 all apply
    the only one that you could think might not apply would be iii) but since m is constant it reduces exactly to ii)

    for question 3
    the gradient of the potential energy points UP the slopes, the forces push you down the slopes so F=-Grad(U)
     
  4. Dec 17, 2011 #3

    gneill

    User Avatar

    Staff: Mentor

    Question 1 could be a trick question -- Note that all the values on the RHS of the equation are vector quantities, while the F's on the LHS are scalar in every case. That makes them all incorrect.
     
  5. Dec 17, 2011 #4
    1) If the mass is constant then i), ii) iii) are all equivalent, so i guess you could use them all.

    The question itself states which of the following forms of N2L may be used, well if it is a trick then the question itself has stated N2L incorrectly, so it should say which of the equation can be used to find a magnitude of Force, not N2L because N2L has been stated wrongly.

    2) Isn't it W= Fcos(theta) where theta is angle from the ground. so theta = 60

    3) F= -Grad(U) as stated which is why it is negative.
     
  6. Dec 17, 2011 #5
    I'll ask our professor when we go back to Uni about question one, I didn't expect to see a trick question like that but it could be. I would have guessed the answer was them all.

    For two, if the answer is 60° then I must have labelled the triangle wrong, that would mean the hyp: 40, adj:20. I had it as hyp:40, opp:20. I'll go try it again.

    And for three, thank you everyone for clearing that up.

    I'll post back here if I have any more problems with other questions I come across.
     
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