Question

A car begins to move at time $t=0$ and then accelerates along a straight track with a speed given by $V\left(t\right)=2t^{2}m{s}^{-1}$ for $0\le t\le 2$.
After the end of acceleration, the car continues to move at constant speed. A small block initially at rest on the floor of the car begins to slip at $t=1sec$ and stops slipping at $t=3sec$. Find the coefficient of static and kinetic friction between the block and the floor.

• A. ${\mu }_s=0.4,{\mu }_k=0.3$
• B. ${\mu }_s=0.3,{\mu }_k=0.4$
• C. ${\mu }_s=0.2,{\mu }_k=0.5$
• D. ${\mu }_s=0.5,{\mu }_k=0.2$

Answer 1

Answer: (A)

${\mu }_s=0.4,{\mu }_k=0.3$

$a\left(t\right)=\frac{dv\left(t\right)}{dt}=4t$
At $t=1$, $a=4m/s^2$
When the block is just starting to slip we have,
${\mu }_{s}N=ma$
$\Rightarrow {\mu }_{s}mg=4m$
$\Rightarrow {\mu }_s=0.4$
Now it goes on slipping till 3 seconds, therefore in 2 seconds the block's velocity changes from $2to2\times (2{)}^2=8m/sec$
Hence, acceleration from $t=1$ to $t=3$ secs is: $a_{avg}=\frac{8-2}{2}=3m/s^2$
Applying equation of motion
${\mu }_k\times m\times g=m\times a_{avg}=3m$
$\Rightarrow {\mu }_k=0.3$