Question

An object is gently placed on a long converges 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

$15.125m$

Step1: Given data.

Initial velocity, $u=11m{s}^{-1}$

Coefficient of friction, $\mu =0.4$

Step2: Find the distance covered by the block.

Formula used:

$v^2-u^2=2as$

Where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration, $s$ is distance.

$f_r=\mu mg$

Where $f_r$ is the frictional force, $m$ is the mass of the block, $g$ is the acceleration due to gravity.

Since,

We know that, retardation due to friction.

$\Rightarrow$$a=\mu g$ $\left[\because f_r=m\alpha \right]$

$\Rightarrow$$a=0.4\times 9.8$ $\left[\because g=9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right]$

$\Rightarrow$$a=3.92m{s}^{-2}$ …..$\left(i\right)$

Since this is retardation therefore there will be a negative sign.

$a=-3.92m{s}^{-2}$

Now,

$v^2-u^2=2as$

$\Rightarrow$$0^2-{11}^2=2\times \left(-3.92\right)\times s$

$\Rightarrow$ $s=\raisebox{1ex}{$-121$}\!\left/ \!\raisebox{-1ex}{$-8$}\right.$

$\Rightarrow$ $s=15.433m\approx 15.125m$

Hence, option B is correct.