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 30.0 m/s

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

${\mathrm{F}}_{\mathrm{f}}={\mathrm{μ}}_{\mathrm{s}}\mathrm{mg}$

Here, ${\mathrm{μ}}_{\mathrm{s}}$is the coefficient of static friction, m is the mass of the tool box, and g is the acceleration due to gravity.

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

$$\begin{gathered} {\mathrm{F}}_{\mathrm{f}}=0.650×\mathrm{m}×\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =\left(6.37×\mathrm{m}\right)\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The expression for maximum possible acceleration can be expressed as,

$\mathrm{a}=\frac{{\mathrm{F}}_{\mathrm{f}}}{\mathrm{m}}$

Substitute role="math" localid="1667644013907" $\left(6.37×\mathrm{m}\right)\mathrm{m}/{\mathrm{s}}^2$for ${\mathrm{F}}_{\mathrm{f}}$ in the above equation.

$$\begin{gathered} \mathrm{a}=\frac{\left(6.37×\mathrm{m}\right)\mathrm{m}/{\mathrm{s}}^2}{\mathrm{m}} \\ =6.37\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the acceleration is $6.37\mathrm{m}/{\mathrm{s}}^2$.

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

v = u + a t

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

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

$$\begin{gathered} 30.0\mathrm{m}/\mathrm{s}=0\mathrm{m}/\mathrm{s}+\left(6.37\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t} \\ \mathrm{t}=\frac{30.0\mathrm{m}/\mathrm{s}}{6.37\mathrm{m}/{\mathrm{s}}^2} \\ =4.7\mathrm{s} \end{gathered}$$

Hence, required shortest time is, 4.7 s.