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How to find acceleration from force and time?

  1. Nov 8, 2016 #1
    1. The problem statement, all variables and given/known data
    A coin is dropped into a well and the time taken to hit the bottom is 2.48 seconds. Calculate the mass of the coin if the water slows it to a stop in 0.72 seconds exerting an average force of 0.41N on the coin

    2. Relevant equations
    F=ma m=f/a

    3. The attempt at a solution
    Finding the mass from m=f/a but first finding the acceleration with just the time and force values
     
  2. jcsd
  3. Nov 8, 2016 #2

    BvU

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    Hi day, :welcome:

    It is clear you'll need another relevant equation: in yours there is no time variable present. Any idea what to do with the 2.48 seconds, for example ?
     
  4. Nov 8, 2016 #3
    Hmm I didn't think the 2.48s was relevant to the mass but it very well could be. The other two parts of the question before this ask for the depth of the well and the final velocity of the coin - which I derived from the suvat equation s=ut+1/2at^2 which I rearranged for the acceleration of 0.806m/s^2. I also got the displacement of the fall as 4.959m and the final velocity as 1.999m. These are probably relevant to the final question and the mass but i'm not sure how..
     
  5. Nov 8, 2016 #4
    Ahh, I think I've forgotten to include gravity 9.81ms^2 in the original equation as acceleration.
     
  6. Nov 8, 2016 #5
    Can you find the deceleration caused by the water ?
    and then use ##f = ma##.
     
  7. Nov 8, 2016 #6
    Possibly. I'm not sure how to calculate acceleration in water - would the 0.41N of force from the water be a clue? Once I've got that I can find a new velocity with the 0.72s. I may have to just use the final velocity from the fall and carry it over? It is just 0.72s afterall...
     
  8. Nov 8, 2016 #7

    gneill

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    Draw an FBD for the coin in water. What are the forces acting and in which directions?
     
  9. Nov 8, 2016 #8
    try ##v-u=at## to find declaration.
     
  10. Nov 9, 2016 #9

    BvU

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    ##{1/2}\;gt^2## (an expression that does include a time!) gives me something else for ##t=2.48## s. And 1.999 m is not a speed. Could you show the complete exercise text and post your working (instead of some answer numbers)?
     
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