The time taken by a block of mass m to slide | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $L$ be the length of the incline.For a smooth plane, the acceleration is $a_1 = g \sin	heta$. The time taken is $t_1 = \sqrt{2L/a_1}$.For a r

Vedclass · Accepted Answer

$5$/$9$

Vedclass · Answer

(C) Let $L$ be the length of the incline.For a smooth plane, the acceleration is $a_1 = g \sin	heta$. The time taken is $t_1 = \sqrt{2L/a_1}$.For a rough plane, the acceleration is $a_2 = g(\sin	heta - \mu \cos	heta)$. The time taken is $t_2 = \sqrt{2L/a_2}$.Given that $t_2 = 1.5 t_1$, we have $t_2^2 = 2.25 t_1^2$.Substituting the expressions for $t_1$ and $t_2$, we get $\frac{2L}{a_2} = 2.25 \frac{2L}{a_1}$, which simplifies to $a_1 = 2.25 a_2$.Substituting $a_1$ and $a_2$, we get $g \sin	heta = 2.25 g(\sin	heta - \mu \cos	heta)$.Since $	heta = 45^\circ$, $\sin	heta = \cos	heta = 1/\sqrt{2}$.Dividing by $g \sin	heta$, we get $1 = 2.25(1 - \mu)$.Thus, $1 - \mu = 1/2.25 = 4/9$.Therefore, $\mu = 1 - 4/9 = 5/9$.

Similar Questions

$A$ small block starts sliding down an inclined plane forming an angle $45^{\circ}$ with the horizontal. The coefficient of friction $\mu$ varies with distance $s$ as $\mu = C s^2$, where $C$ is a constant of appropriate dimensions. The distance covered by the block before it stops is:

$A$ given object takes $n$ times more time to slide down a $45^o$ rough inclined plane as it takes to slide down a perfectly smooth $45^o$ incline. The coefficient of kinetic friction between the object and the incline is

When a body is placed on a rough plane (coefficient of friction $= \mu$) inclined at an angle $\theta$ to the horizontal,its acceleration is (acceleration due to gravity $= g$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module