# Thread: Is this question solvable?

1. ## 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?

2. 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

3. 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.

4. 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.