Results 1 to 4 of 4

Math Help - Is this question solvable?

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    86

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

    Required:
    m = ?

    Analysis:

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

    One of the two values has now been obtained to solve for m.
    \frac{F_net}{a} = m
    The F_net value is still needed. To find this value, plug the corresponding values into the equation F_net = F_App + F_fk. To find F_fk, plug the corresponding values into the equation F_N*\mu_k = F_fk. The F_N value is still needed. This value is equal to F_G. To find this value, plug the corresponding values into the equation 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; April 10th 2010 at 12:36 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244
    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:

    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)

    ma+\mu_kmg=F_{App}

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

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

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2009
    Posts
    86
    Thanks for the reply. I'm unsure that F_{net} = F_{app} -  F_f, as my textbook and teacher says that F_{net} = F_{app}  + F_f. I manage to rearrange the equation m*a = F_{app} +  \mu_k*m*g into  m = \frac{F_{app}}{a - \mu_k*g}. I then solved it as follows:

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

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

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

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

    m = 1358.23kg

    The text book answer 1400kg rounded. So I think this answer is accurate. Correct me if I'm wrong.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. S4 solvable
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: May 5th 2011, 11:46 PM
  2. Solvable Group
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 12th 2010, 01:14 PM
  3. solvable?
    Posted in the Business Math Forum
    Replies: 7
    Last Post: November 5th 2009, 08:54 PM
  4. Is this equation solvable?
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: July 22nd 2009, 07:31 PM
  5. Is this solvable?
    Posted in the Algebra Forum
    Replies: 0
    Last Post: March 11th 2009, 10:00 PM

Search Tags


/mathhelpforum @mathhelpforum