A 12.0 kg box rests on the level bed of a truck. The coefficients of friction between the box and bed are ${\mathbf{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{19}\mathbf{\text{and}}{\mathbf{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{15}$. The truck stops at a stop sign and then starts to move with an acceleration of 2.20$\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ . If the box is from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck? How far does the truck travel in this time?

The given data can be listed below as:

• The mass of the box is m = 12 kg .
• The coefficient of the static friction is${\mu }_s=0.19$ .
• The coefficient of the kinetic friction is${\mu }_k=0.15$.
• The acceleration of the truck is$a=2.20m/s^2$.
• The distance of the box from the truck is d = 1.80 m .

Newton’s second law states that the force exerted on an object is equal to the product of the mass and the acceleration of that object. The force exerted is equal to the rate of change of momentum of that object.

The free-body diagram of the box has been drawn below:

Here, the normal force N is acting upwards and the weight of the box mg is acting in the downwards direction. The frictional force f is also acting in the left direction.

As the box moves in the direction opposite to that of the truck.

F = -ma

Here, F is the force exerted, m is the mass of the box and a is the acceleration of the box. The net force is negative.

The equation of the frictional force is expressed as:

-f = - ma …(i)

Here, f is the frictional force.

The equation of the frictional force is expressed as:

$f={\mu }_{k}mg$

Here, ${\mu }_k$is the kinetic friction and is the acceleration due to gravity.

Substitute values in the above equation-

$\begin{array}{r}{\mu }_{k}mg=ma\\ a={\mu }_{k}g\end{array}$

Here, a is the acceleration of the box.

Substitute the values in the above equation.

$\begin{array}{r}a=\left(0.15\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\\ =1.47\mathrm{m}/{\mathrm{s}}^2\end{array}$

The equation of the time required to travel the distance is expressed as:

$d=ut+\frac{1}{2}a{t}^2$

Here, d is the distance traveled by the, u is the initial velocity of the truck and t is the time required to travel the distance.

As the truck was at rest initially, the initial velocity of the truck is zero.

Substitute the values in the above equation.

$\begin{array}{r}1.80\mathrm{m}=\left(0\right)\mathrm{t}+\frac{1}{2}\left(1.47\mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2\\ 1.80\mathrm{m}=\left(0.735\mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2\\ {\mathrm{t}}^2=2.44{\mathrm{s}}^2\\ \mathrm{t}=1.57\mathrm{s}\end{array}$

Thus, the box falls off the truck in1.57 s .

The equation of the distance travelled is expressed as:

$d=ut+\frac{1}{2}a{t}^2$

Here, d is the distance travelled by the truck.

As the truck was at rest initially, the initial velocity of the truck is zero.

Substitute the values in the above equation.

$\begin{array}{r}\mathrm{d}=\left(0\right)(1.57\mathrm{s})+\frac{1}{2}\left(1.47\mathrm{m}/{\mathrm{s}}^2\right)(1.57\mathrm{s}{)}^2\\ =\left(0.735\mathrm{m}/{\mathrm{s}}^2\right)\left(2.4{\mathrm{s}}^2\right)\\ =1.81\mathrm{m}\end{array}$

Thus, the truck has travelled 1.81 m.