In the school cafeteria, a trouble-making child blows a 12.0 g spitball through a 25.0 cm straw. The force
of her breath as a function of the distance along the length of the straw can be modeled as 10-50x^2-70x^3
where force is in newtons and x is in meters. (a) How much work is done by her breath on the spitball as it travels the length of the barrel? (b) Assuming negligible friction and the straw is held horizontally, what is the speed of the spitball as it leaves the straw?

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  1. F = 10 - 50x^2 - 70x^3.
    F=10 - 50*(0.25)^2 - 70*(0.25)^3=5.78 N.

    a. W = F * d = 5.78 * 0.25=1.45 Joules.

    b. a = F/m = 5.78 / 0.012 = 481.7 m/s^2.
    V^2 = Vo + 2a*d.
    V^2 = 0 + 963.4*0.25 = 240.85
    V = 15.5 m/s.

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  2. This is incorrect on part A. Work is a measurement of force over position / time, and since the force of the spitwad as it moves through the straw is a function of it's position, the force at the initial position 0 is 10 Newtons, and then ~5 - 25/32 Newtons or so at 0.25 meters. To find work, integrate the function of the force from position 0 to position 0.25.

    Integral (10-50*x^2-70*x^3, x, 0, 0.25)

    The area under the force curve gives you the work done for all changing points from 0 to 0.25.

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  3. As a note, multiplying the force by the change in distance is only applicable if the force is constant. Multiplying a constant linear force by the change in distance gives you the area under the constant force curve.

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