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A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed 14.0 m/s, the total force exerted on the driver has magnitude 115 N. What are the magnitude and direction of the total vector force exerted on the driver if the speed is 25.0 m/s instead?

I thought you would use proportions, but its not right. THANK YOU

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4 answers

  1. Ignoring gravity

    v^2/R = centripetal acceleration

    So force toward center = mv^2/R
    115 N = m 14^2/R
    so
    m/R = 115/14^2

    F = m 25^2/R = (115/14^2) (25^2)

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  2. Thank you so much! Makes perfect sense =D

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  3. You are welcome :)

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  4. In ∑F=m
    r
    v
    2


    , both m and r are unknown but remain constant.
    Symbolically, write
    ∑F
    slow

    =(
    r
    m

    )(14.0m/s)
    2
    and ∑F
    fast

    =(
    r
    m

    )(18.0m/s)
    2

    Therefore, ∑F is proportional to v
    2
    and increases by a factor of (
    14.0
    18.0

    )
    2
    as v increases from 14.0m/s to 18.0m/s. The total force at the higher speed is then
    ∑F
    fast

    =(
    14.0
    18.0

    )
    2
    ∑F
    slow

    =(
    14.0
    18.0

    )
    2
    (130N)=215N

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