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Thread: Is this question solvable?

  1. #1
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    Is this question solvable?

    During a "Spirit Week" rally, twenty students, each exerting a force of 8000N [forward], pushed a teacher's car on level ground and accelerated it from rest to 1.5m/s [forward] in 3.0s. If the coefficient of friction was approximately 0.55 for both the static and kinetic cases, determine the mass of the teacher's car.

    Given:
    $\displaystyle F_App = 8000N[forward]$
    $\displaystyle v_1 = 0m/s$
    $\displaystyle v_2 = 1.5m/s[forward]$
    $\displaystyle t = 3s$
    $\displaystyle \mu_k = 0.55$
    $\displaystyle \mu_s = 0.55$

    Required:
    m = ?

    Analysis:

    First solve for acceleration by plugging the given values into the equation $\displaystyle \frac{v_2 - v_1}{t} = a$
    $\displaystyle \frac{1.5m/s[forward] - 0m/s}{3s} = a$
    $\displaystyle 0.5m/s^2[forward] = a$

    One of the two values has now been obtained to solve for $\displaystyle m$.
    $\displaystyle \frac{F_net}{a} = m$
    The $\displaystyle F_net$ value is still needed. To find this value, plug the corresponding values into the equation $\displaystyle F_net = F_App + F_fk$. To find $\displaystyle F_fk$, plug the corresponding values into the equation $\displaystyle F_N*\mu_k = F_fk$. The $\displaystyle F_N$ value is still needed. This value is equal to $\displaystyle F_G$. To find this value, plug the corresponding values into the equation $\displaystyle m*g = F_G$.

    This is where the problem is. I need the mass in order to solve for the mass. Does anyone know any other way of solving this? Or was the question erroneously written to a point where it can't be solved?
    Last edited by shadow6; Apr 10th 2010 at 11:36 AM.
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  2. #2
    MHF Contributor
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    I think you have everything correct, except for the net force is equal to the applied force *minus* the frictional force. Once you get everything into one equation, you can solve for m:

    $\displaystyle m=\frac{F_{net}}{a}=\frac{1}{a}(F_{App}-F_{fk})=\frac{1}{a}(F_{App}-\mu_kF_N)=\frac{1}{a}(F_{App}-\mu_kmg)$

    $\displaystyle ma+\mu_kmg=F_{App}$

    $\displaystyle m=\frac{F_{App}}{a+\mu_kg}$

    Post again in this thread if you're still having trouble.

    - Hollywood
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  3. #3
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    Thanks for the reply. I'm unsure that $\displaystyle F_{net} = F_{app} - F_f$, as my textbook and teacher says that $\displaystyle F_{net} = F_{app} + F_f$. I manage to rearrange the equation $\displaystyle m*a = F_{app} + \mu_k*m*g$ into $\displaystyle m = \frac{F_{app}}{a - \mu_k*g}$. I then solved it as follows:

    $\displaystyle m = \frac{8000N[forward]}{0.5m/s^2[forward] - 0.55*9.8m/s^2}$

    $\displaystyle m = \frac{8000N[forward]}{0.5m/s^2[forward] - 5.39m/s^2[backwards]}$

    $\displaystyle m = \frac{8000N[forward]}{0.5m/s^2[forward] + 5.39m/s^2[forward]}$

    $\displaystyle m = \frac{8000N[forward]}{5.89m/s^2[forward]}$

    $\displaystyle m = 1358.23kg$

    The text book answer 1400kg rounded. So I think this answer is accurate. Correct me if I'm wrong.
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  4. #4
    MHF Contributor
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    I think we're just saying the same thing in different ways. I defined the frictional force opposite to the applied force and subtracted a positive value; you defined the frictional force in the same direction as the applied force and added a negative value.

    Your solution is correct.

    - Hollywood
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