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At what speed does the food enter the stomach

  1. Aug 16, 2005 #1
    A morsel of food with a mass of 4.2g is injected into the esophagus with an initial speed of 2.5cm/s. On the way down to the stomach, the walls of the esophagus exert an upward resistive force of 0.0027 N on the morsel. Ifthe esophagus is 20cm long, with what speed does the morsel of food enter the stomach?

    I solve it by :

    Wf = 1/2 m (v^2 - u^2) and I got the wrong answer.

    Can anyone tell me what`s wrong with it?

    I feel very blur with enerygy concept, especially relate with non conservative force. Can anyone summarize the important concept about energy for me? :frown:
  2. jcsd
  3. Aug 16, 2005 #2


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    I don't see that energy has anything to do with it.

    Although you say "down" the gullet, I feel sure you are not assuming the esophagus is not straight up and down and that there is no gravitational acceleration downward!
    In that case, the only force on the food is the upward force of 0.0027 N and so the acceleration is 0.0027N/0.0042 kg= 0.64 m/s2= 64.29 cm/s2.

    Now use d= (1/2)at2+ v0t and v= at+ v0
    with a= 64.29 dm/s2, v0= -2.5 c/s, and d= -20. Solve the first equation for t, the time the food takes to get to the stomach, then use the second equation to find v, the speed of the food at that time.

    Having said all of that, I'm going to rethink what I initially said: You can use the mass and initial speed of the food to find its initial kinetic energy. That is decreased by the work the esophagus does on the food: Work= Force*distance= 0.0027 N* 0.20 m. Subtracting that from the initial kinetic energy gives the final kinetic energy of the food: you can find the speed from that.

    Be sure to use consistent units! Either meters, kilograms, Newtons, and Joules or cm, grams, dynes, and ergs.
  4. Aug 16, 2005 #3
    I get this equation from my textbook:

    Work done by non conservative force = change in KE + change in PE

    If I apply this formula in this question, why cant i find the true answer?
  5. Aug 16, 2005 #4


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    It appears that you are not using the CHANGE in KE and PE in your original equation. Since the food has an initial velocity and a final velocity, there need to be two terms for the velocity in the KE portion of the equation.

    Start by calculating each component of your equation:
    WORK: [tex]W_f = F*d[/tex]

    KE: [tex]\Delta KE = \frac{1}{2} m ({v_2}^2 - {v_1}^2)[/tex]

    PE: [tex]\Delta PE = m*g (h_2 - h_1)[/tex]

    You know every component except [tex]v_2[/tex]
    Last edited: Aug 16, 2005
  6. Aug 16, 2005 #5
    I got it. Thanks.
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