Question

If a block moving up an inclined plane at ${30}^o$ with a velocity of $5m/s$, stops after $0.5s$, then coefficient of friction will be nearly

• A. $0.5$
• B. $0.6$
• C. $0.9$
• D. $1.1$

Answer 1

The correct option is B $0.6$

We know that ${}_V=u+(a\times t)$

Here ${}_V=0,u=5m/s$ and $a$ is net acceleration

$\therefore 0=5-(a\times 0.5)$ the minus sign indicates that it is retardation

$\therefore a=10m/s^2$ a is total retardation.

Here the total retardation is due to friction and gravity.

The diagram shows the situation. Gravity $mg$ is vertically downwards. Its component which opposes its motion is $mg\times cos\left({60}^o\right)=\frac{1}{2}mg.$

The friction is due to normal reaction which is always perpendicularly upwards the surface. This reaction is the opposition force to the component of gravity which is in the line of $N$ but directed downwards. That component is $mg\times cos\left({30}^o\right)=\frac{\sqrt{3}}{2}mg.$

So the net force of friction is $\mu \times \frac{\sqrt{3}}{2}mg=\frac{\sqrt{3}}{2}\mu mg$

Hence total retarding force is $ma=\frac{1}{2}mg+\frac{\sqrt{3}}{2}\mu mg$

cancelling $m$ we get

$a=\frac{g}{2}+\frac{\sqrt{3}\mu g}{2}=\frac{g}{2}(1+\sqrt{3}\mu )$

$\therefore \frac{2a}{g}=1+\sqrt{3}\mu$

$\therefore \frac{2a-g}{\sqrt{3}g}=\mu$

$\therefore \mu =\frac{(2\times 10)-9.8}{1.732\times 9.8}=0.6$

$\mu$ has no unit.