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Work from acceleration vs time

  1. Oct 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Given Data: Graph, 2 kg particle.


    I am asked to calculate the work done from 0 to 4 meters, or some other interval. Normally I would simply multiply the FORCE of the given particle by multiplying 6m/s^2 times 2kg to create a new force vs. displacement graph, and then calculate the area under the linear piecewise curve. I am doing something wrong


    2. Relevant equations

    acceleration_vs_time_work_by_falchiongpx-d82wk1e.jpg

    3. The attempt at a solution

    Calculate the forces as 2kg*acceleration to get the variable force y-values. Calculate area under curve... does not work according to book. Do not seem have enough info for an integral solution.

    Sample attempt for Work (J) done from 1 to 2 meters..

    1/2(1m*(2kg*18 m/s^2))
     
  2. jcsd
  3. Oct 15, 2014 #2

    Simon Bridge

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    You appear to have an acceleration vs displacement graph (bottom horizontal axis).
    Work is the area under the force vs displacement graph.
    From 1-2m the area is a triangle ... which you did.
    What is the problem?
     
  4. Oct 15, 2014 #3

    NTW

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    In the first meter, judging from the picture, the triangle under the line has an area of (1 m*18 m/s2)/2 = 9 m2/s2

    Now you multiply that by the 2kg of mass, and get 18 Joule. The units are right: acceleration [LT-2] by displacement [L] by mass [M], hence ML2T-2.

    That way, you can continue the integration 'by eye'...
     
  5. Oct 15, 2014 #4
    I found out that I misreading the graph from the book. The book labels the scale as = 6m/s2. Every horizontal line represented was actually 2m/s2, with the graph maxing out at 6m/s2. Unfortunately, as was actually the MAX value for the graph. My problem was with reading the scale incorrectly. The teacher was kind enough to go over the problem with me where I realized that each increment of the y-value was 2 m/s2, instead of 6 m/s2.

    Note: apologies if I mislabeled the last graph. Please let me know if this is wrong.

    Corrected Graph Accel vs Displacement Graph, this is a better representation of what was in the book:

    While as = 6m/s

    acceleration_vs_disp_work_corrected_by_falchiongpx-d82x92j.jpg

    Solution:
    a) Read the graph carefully, and locate all notations, even if you require a microscope.
    b) Convert y-values on the graph from acceleration to FORCE via F=ma:
    6m/s2(2 kg)=12 Newtons, hold the figs

    i. Re-graph as a graph of F (x), and calculate the area under the curve for the W.


    f_x__vs_disp_for_work_by_falchiongpx-d82x9f4.jpg

    ii. For finding the velocity at x meters, take the integral of F(x)...

    Antiderivative of F(x) results in 1/2mv2
    InteGRAL over a an interVAL.
    (I'm new to integration)

    W=W=1/2mv2 - 1/2mv20

    Which is coincidental:
    ΔKE=KE-KE0

    And where initial velocity is zero.

    v=√[(2*J)/(m)] → v=√[(2*J)/(2 kg)]
    Does the particle change direction? Answer: what has the greatest energy? The energy for a slowdown or the energy for a speedup?

    This allows for solving Work Done, and Velocity from a(x). Acceleration as it relates to Displacement.​
     
    Last edited: Oct 15, 2014
  6. Oct 15, 2014 #5

    Simon Bridge

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    ... or use the work-energy theorem.
     
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