Question

A body is moving down along an inclined plane of slope ${37}^{\circ }$. The co-efficient of friction between the body and plane varies as $\mu =0.3x$, where $x$ is distance travelled down the plane. The body will have maximum speed at $x$ equal to
$[\sin {37}^{\circ }=3/5\text{and}g=10{\text{m/s}}^2]$

• A. $2.5\text{m}$
• B. $2.26\text{m}$
• C. $1.16\text{m}$
• D. $2\text{m}$

Answer 1

The correct option is A $2.5\text{m}$


From FBD of block,

$N=mg\cos {37}^{\circ }$

$f=\mu N=\mu mg\cos {37}^{\circ }$

From Newton's second law,

$mg\sin {37}^{\circ }-f=ma$

$\Rightarrow mg\sin {37}^{\circ }-0.3xmg\cos {37}^{\circ }=ma$

The body will have maximum speed when net force on it is zero or $ma=0$

$\therefore mg\sin {37}^{\circ }-0.3x\times mg\cos {37}^{\circ }=0$

$\Rightarrow x=\frac{mg\sin {37}^{\circ }}{0.3\times mg\cos {37}^{\circ }}$

$\Rightarrow x=\frac{\left(3/5\right)}{0.3\times \left(4/5\right)}=2.5\text{m}$

Hence, option $\left(d\right)$ is correct.

$$\begin{array}{c}\text{Why this question ?}\\ \text{Concept-A body have maximum velocity}\\ \text{in a particular direction when net force on it is}\\ \text{zero. If opposing force is increased more then}\\ \text{it will start decelerating and come to rest}\\ \text{at a point and then start accelerating in the}\\ \text{direction of greater force.}\end{array}$$