Question

An object is gently placed on a long conveyor belt moving with $11m{s}^{-1}$. If the coefficient of friction is $0.4$, then the block will slide in the belt up to a distance of

• A. $10.21m$
• B. $15.125m$
• C. $20.3m$
• D. $25.6m$

Answer 1

The correct option is B

$15.125m$

Step 1. Given data:

The velocity of the conveyor belt, $u^{\text{'}}=11m{s}^{-1}$

Coefficient of static friction, $\mu =0.4{}_{}$

Initially velocity of the object with respect to the belt, $u=11m{s}^{-1}$

Frictional force, $F=\mu mg$

Step 2. Find acceleration,$a$

From free body diagram, $F=ma$

$\Rightarrow$ $\mu mg=ma$

$\Rightarrow$ $$\begin{gathered} a=\mu g \\ =0.4x10 \\ =4m{s}^{-2} \end{gathered}$$ [$g=10m{s}^{-2}$]

In the frame of the belt, the final velocity of the object with respect to the belt will be $v=0$.

Step 3. Find the distance:

Using the equation, $v^2=u^2+2as$

$\Rightarrow$ $0={\left(11\right)}^2-2\left(4\right)s$ [$a=-4m{s}^{-2}$] Here the negative sign of acceleration indicates that the velocity of the object is decreasing, thus, the object is decelerating.

$\Rightarrow$ $0=121-8s$

$\Rightarrow$ $s=\frac{121}{8}$

$\Rightarrow$ $s=15.125m$

The distance of the object with respect to the ground is$\mathbf{15}\mathbf{.}\mathbf{125}\mathbf{}\mathit{m}$.

Hence, option (B) is the correct option.