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Finding Work Done, Acceleration, Force and Time

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data
    An object with a mass of 2kg gets accelerated from 15m/s to 25m/s.

    A: What is the work done in effecting this velocity change?
    B: The distance moved was 8m, what force produced the acceleration?
    C: How long did it take to produce the acceleration (presumed uniform)?

    2. Relevant equations
    Equations of Motion

    3. The attempt at a solution
    Part A:
    W=\frac{1}{2}m(v^2 - u^2) \\
    W=\frac{1}{2}2(25^2 - 15^2) \\
    W= 25^2 - 15^2 = 400J

    Part B:
    v^2=u^2+2as \\
    ∴ a = \frac{v^2 - u^2}{2s} \\
    F=ma \\
    F=2 \times \frac{25^2 - 15^2}{2 \times 8} \\
    F=2 \times \frac{400}{16} \\
    F=2 \times 25 = 50N

    Part C
    v=u+at \\
    ∴ t = \frac{v-u}{a} \\
    t= \frac{25-15}{25} = 0.4s

    My biggest problem is the values, they just dont seem right to me, 0.4s to accelerate a 2kg mass from 15 to 25m/s using 50N over 8m?

    Any help appreciated :).
  2. jcsd
  3. Apr 16, 2013 #2
    look at the definition of work, for part b.
  4. Apr 16, 2013 #3
    For part (c), you can use the impulse-momentum theorem to check your answer
  5. Apr 16, 2013 #4
    I keep reading it but cant seem to find another way of interpreting it. Surely it means what force produced the acceleration (which is 25m/s^2 from my calculations) and then that comes out to be 50N right?

    Any little hint at just interpreting it would be appreciated :) .
  6. Apr 16, 2013 #5
    Are right, is that that the impulse is equal to the change in momentum?

    that would be then
    I=Ft \\
    I=50 \times 0.4 = 20Ns \\

    Δp=mv-mu \\
    Δp=2 \times 25 - 2 \times 15 = 50-30 = 20Ns

    Does that mean it is correct?
  7. Apr 16, 2013 #6
    It looks right :)
    It's always useful if you can find another way of doing a problem to check your answers.

    For part (b) you can do a similar answer check by using the definition of work (as mentioned by Clever_name).
  8. Apr 16, 2013 #7
    Ah right, I see now! force times distance, I thought clever_name meant for me to double check what the question was asking lol.
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