Question

The backside of a truck is open and a box of $40kg$ is placed $5m$away from the rear end. The coefficient of friction of the box with the surface of the truck is $0·15$ The truck starts from rest with $2m/s^2$ acceleration. Calculate the distance covered by the truck when the box falls off

• A. $20m$
• B. $30m$
• C. $40m$
• D. $50m$

Answer 1

The correct option is A

$20m$

Step1: Given data

Mass of box$\left(m\right)=40kg$

Distance between box and rear end $\left(s\right)=5m$

Coefficient of friction of the box with the surface of the truck$\left(\mu \right)=0·15$

Acceleration of truck$\left(a\right)=2m/s^2$

Considering acceleration due to gravity $g=10m/s^2$

Step 2: Formula used

To determine the acting force using the formula, $Force\left(F\right)=mass\left(m\right)\times acceleration\left(a\right)$

The second equation of motion $S=ut+\frac{1}{2}a{t}^2$

Where, $u$ is the initial velocity, $a$ is acceleration

Step 3: Compute the net force

Substitute the known values in the formula, $Force\left(F\right)=mass\left(m\right)\times acceleration\left(a\right)$

$$\begin{gathered} F=40\times 2 \\ F=80N \end{gathered}$$

But the frictional force is also acting on the box.

We know that,

The normal reaction is equal to weight in a downward direction.

$N=mg$

$$\begin{gathered} Frictionalforce\left(F_f\right)=coefficientoffriction\left(\mu \right)\times normalforce\left(N\right) \\ {F}_f=\mu \times mg \\ {F}_f=0·15\times 40\times 10 \\ {F}_f=60N \end{gathered}$$

So, the net acting force is,

$$\begin{gathered} F-F_f=netforce \\ 80-60=20N \end{gathered}$$

Step4: Obtain the backward acceleration

The produce backward acceleration is,

$$\begin{gathered} a_b=\frac{F_{net}}{m} \\ {a}_b=\frac{20}{40} \\ {a}_b=0·5m/s^2 \end{gathered}$$

Step4: Determine the time taken to travel $\left(s\right)=5m$

Substitute the known values in the second equation of motion $S=ut+\frac{1}{2}{a}_b{t}^2$ to obtain the time,

$$\begin{gathered} 5=0\times t+\frac{1}{2}\times 0·5t^2 \\ 10=0·5t^2 \\ {t}^2=\frac{10}{0·5} \\ {t}^2=20 \\ t=\sqrt{20}sec \end{gathered}$$

So, the distance $\left(s\right)=5m$ is traveled by truck in $\sqrt{20}sec$.

Step 5: Determine the distance traveled by truck

Suppose the distance covered by the truck is $x$. Since initially it is at rest so $u=0$

$$\begin{gathered} S=ut+\frac{1}{2}a{t}^2 \\ x=0\times t+\frac{1}{2}\times 2\times {\left(\sqrt{20}\right)}^2 \\ x=20m \end{gathered}$$

So, the distance covered by the truck when the box falls off is $20m$.

Hence, option A is the correct answer.