Question 23: Laws of Motion

Question

A cube bar slides thrice the time with friction than that without friction when it slides on an inclined plane of inclination $$45^{\circ}$$. The coefficient of friction between the block and the surface is

Accepted Answer

To solve this problem, we need to compare the time taken for the block to slide down the inclined plane with and without friction.
Without Friction:
When there is no friction, the only force acting along the inclined plane is the component of the gravitational force. The acceleration $a_{	ext{nf}}$ (where "nf" stands for no friction) can be given by:
$$ a_{	ext{nf}} = g \sin(	heta) $$
Where:
$g$ is the acceleration due to gravity.
$	heta = 45^\circ$, so $\sin(45^\circ) = \frac{\sqrt{2}}{2}$.
Thus:
$$ a_{	ext{nf}} = g \frac{\sqrt{2}}{2} $$
With Friction:
When friction is present, the net force along the plane is reduced by the frictional force. The acceleration $a_{	ext{f}}$ (where "f" stands for friction) is:
$$ a_{	ext{f}} = g \sin(	heta) - \mu g \cos(	heta) $$
Where:
$\mu$ is the coefficient of friction.
$\cos(45^\circ) = \frac{\sqrt{2}}{2}$.
Thus:
$$ a_{	ext{f}} = g \frac{\sqrt{2}}{2} - \mu g \frac{\sqrt{2}}{2} $$
$$ a_{	ext{f}} = g \frac{\sqrt{2}}{2} (1 - \mu) $$
Relation between Times:
From the problem, the time with friction $t_{	ext{f}}$ is three times the time without friction $t_{	ext{nf}}$. Thus:
$$ t_{	ext{f}} = 3 t_{	ext{nf}} $$
Since time $t$ under constant acceleration is inversely proportional to the square root of the acceleration:
$$ t \propto \frac{1}{\sqrt{a}} $$
Therefore:
$$ \frac{1}{\sqrt{a_{	ext{f}}}} = 3 \cdot \frac{1}{\sqrt{a_{	ext{nf}}}} $$
Squaring both sides gives:
$$ \frac{1}{a_{	ext{f}}} = 9 \cdot \frac{1}{a_{	ext{nf}}} $$
Hence:
$$ a_{	ext{f}} = \frac{a_{	ext{nf}}}{9} $$
Substitute for $a_{	ext{f}}$ and $a_{	ext{nf}}$:
$$ g \frac{\sqrt{2}}{2} (1 - \mu) = \frac{g \frac{\sqrt{2}}{2}}{9} $$
Canceling terms common on both sides gives:
$$ 1 - \mu = \frac{1}{9} $$
Solving for $\mu$:
$$ \mu = 1 - \frac{1}{9} $$
$$ \mu = \frac{8}{9} $$
Therefore, the coefficient of friction is approximately 0.88. So, the correct option is:
Option C: 0.88

Answer

1

Answer

0.66

Answer

0.88

Answer

0.5

Laws of Motion

A cube bar slides thrice the time with friction than that without friction when it slides on an inclined plane of inclination $45^{\circ}$. The coefficient of friction between the block and the surface is

Choose the correct answer: