Question

A car begins from rest at time $t=0$ and then accelerates along a straight track during the interval $0\le t\le 2\text{s}$ during which velocity varies as $v=k{t}^2$ . Thereafter it moves with constant velocity as shown in the graph. A coin is initially at rest on the floor of the car. At $t=1\text{s}$, the coin begins to slip and stop slipping at $t=3\text{s}$. The co-efficient of static friction between the floor and the coin is $($ Take $g=10{\text{m/s}}^2)$

• A. $0.2$
• B. $0.3$
• C. $0.4$
• D. $0.5$

Answer 1

The correct option is C $0.4$

Friction force accelarates the coin, when slipping starts.

${\mu }_{s}N=ma$
$\Rightarrow {\mu }_{s}mg=ma$
$\Rightarrow a={\mu }_{s}g$

where, ${\mu }_s\to$ co-efficient of static friction.

Given that graph is a parabola having vertex at origin then function of velocity is $v=k{t}^2$

At $t=2\text{s},V=8\text{m/s}$, thus we have

$8=k\times 2^2$

$\Rightarrow k=2$

$\therefore v=2t^2$

$\Rightarrow a=\frac{dv}{dt}=4t$

The coin begins to slip at $t=1\text{s}$

$\therefore {\mu }_s=\frac{a}{g}=\frac{4\times 1}{10}=0.4$

Hence, option (c) is the correct answer.