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Work on a decelerating car

  1. Oct 18, 2016 #1
    1. The problem statement, all variables and given/known data
    With brakes fully applied, a 1480 kg car deccelerates from a speed of 91.0 km/hr. What is the work done by the braking force in bringing the car to a stop?



    2. Relevant equations
    1/2(mass)(v^2)

    3. The attempt at a solution
    1/2(1480)(25^2) = 462500
    I converted km/hr to m/sec to get 25 for the v value. They want us to include the sign of the acceleration so i added a negative sign to make the answer -462500 but it was wrong (I also tried it without the negative sign). What am I missing? Thank you so much!
     
  2. jcsd
  3. Oct 18, 2016 #2

    TSny

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    Consider the number of significant figures given in the numerical data. When you calculate v in m/s, you should keep at least that same number of significant figures. (I like to keep an extra significant figure during the intermediate calculations and then round off to the appropriate number of significant figures at the end of the calculation.)
     
  4. Oct 18, 2016 #3
    Thanks for your quick reply, TSny, For the significant figures and multiplication I know that you use the value with the least amount of numbers (precision). So for this one would I be basing it off of .5? or 25? If it was .5 I would write it as 4x10^5. If it was 25, I would write it as 4.6 x 10^5. I'm just not sure which one to use? And I would still include the negative sign for the acceleration, do you agree?
     
  5. Oct 18, 2016 #4

    TSny

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    How many significant figures are in the quantity 91.0 km/hr?
    When you convert this to m/s, how many significant figures should you retain?
     
  6. Oct 18, 2016 #5
    Trailing zeros are significant, so 3 sig figs are in 91.0. So converting to m/s it would be 25.0 m/s? So then my answer should be -4.62 x 10^4 retaining the 3 significant figures? Thanks again for being helpful and patient.
     
  7. Oct 18, 2016 #6

    TSny

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    Yes.
    You have the correct number of significant figures, but the last digit is not 0. Check your conversion of km/hr to m/s.
     
  8. Oct 18, 2016 #7
    It looks like you already got your answer. To be explicit, the .5 is a scalar therefore its significance is neglected (see https://en.wikipedia.org/wiki/Significance_arithmetic). I agree with your negative notation since you're going from a higher speed to rest, hence decelerating. (For further clarification, see https://www.av8n.com/physics/acceleration.htm).
    Cheers
     
  9. Oct 18, 2016 #8
    Check your conversion to m/s -- see this for a step-by-step guide.

    Using 25 m/s vs 25.3 m/s makes a huge difference in the answer
     
  10. Oct 18, 2016 #9
    Thanks everyone! I redid my conversion and got 25.3 (rounding up) for my meters per second. Recalculating I got 4.73 x 10^4. Keeping 4.73 consistent with 3 significant figures. And I will mark it as negative since it is decelerating. Did I finally get it?
     
  11. Oct 18, 2016 #10

    TSny

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    The 4.73 looks good, but I think you should check the 104 part.
     
  12. Oct 18, 2016 #11
    That was a silly mistake on my part. Thanks for catching that. My answer in full is 473666.6 which converted into scientific notation will result in a complete answer of -4.73x10^5 or should I round it up to -4.74x10^5? Again, thanks all for the patience.
     
  13. Oct 18, 2016 #12

    TSny

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    If you round 473666.6 to three significant figures, you would "round up" to obtain 474000.
    As I mentioned in an earlier post, I prefer to keep an extra significant figure in intermediate calculations.
    So, I would have obtained v = 25.28 m/s. This would give -472918 for the work, which rounds to -473000 to three significant figures. You can see how you might obtain either -4.73 x 105 or -4.74 x 105. So, in my opinion, either answer should be acceptable. I leave it to you to include the correct units.
     
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