91Ó°ÊÓ

The given data can be listed below.

• The coefficient of kinetic friction is ${\mathrm{μ}}_k=0.355$
• The coefficient of static friction is${\mathrm{μ}}_s=0.650$
• The velocity of the truck is 30.0m/s.

The equation of motion is used for describing the velocity of an object. Using the equation of motion we can easily calculate the acceleration and the time taken by the object.

From the above free-body diagram, the maximum frictional force can be expressed as,

$F_f={\mathrm{μ}}_{s}mg$

Here, ${\mathrm{μ}}_s$is the coefficient of static friction, m is the mass of the toolbox, and g is the acceleration due to gravity.

Substitute 0.650 for ${\mathrm{μ}}_s$,$9.8m/s^2$for g in the above equation.

$\begin{array}{rcl}{F}_f& =& 0.650×m×\left(9.8m/s^2\right)\\ & =& \left(6.37×m\right)m/s^2\end{array}$

The expression for maximum possible acceleration can be expressed as,

$a=\frac{F_f}{m}$

Substitute $\left(\text{6.37}×m\right)m/s^2$for$F_f$in the above equation.

$\begin{array}{rcl}a& =& \frac{\left(\text{6.37}×m\right)m/s^2}{m}\\ & =& 6.37m/s^2\end{array}$

Hence, the acceleration is$6.37m/s^2$.

The expression for first equation of motion can be expressed as,

$v=u+at$

Here, u is the initial velocity, a is the acceleration and t is the time taken.

Substitute 0m/s for u, 30.0m/s for v, and $6.37m/s^2$for a in the above equation.

role="math" localid="1668412426208" $\begin{array}{rcl}30.0m/s& =& 0m/s+\left(6.37m/s^2\right)t\end{array}$

$\begin{array}{rcl}t& =& \frac{30m/s}{6.37m/s^2}\\ & =& 4.7\text{s}\end{array}$

Hence, required shortest time is, 4.7s.