Question

A block of mass $0.5kg$ has an initial velocity of $10m/s$ down an inclined plane ${30}^{\circ }$, the coefficient of friction between the block and the inclined surface is $0.2$. The velocity of the block after it travel a distance of $10m$ figure is:

• A. $13m/s$
• B. $17m/s$
• C. $24m/s$
• D. $8m/s$

Answer 1

The correct option is A $13m/s$

The net force on the block down the incline is:

$F_{net}=mg\sin \theta -\mu N=mg\sin \theta -\mu mg\cos \theta$,

where $m=10kg,g=10m{s}^{-2},\theta ={30}^{\circ },\mu =0.2$ as given.

Thus, $F_{net}\approx 1.63N$, while the acceleration $a$ of the block is given by $a=\frac{F_{net}}{m}\approx 3.27m{s}^{-2}$ down the incline.

Given that $u=10m{s}^{-1},s=10m,$ and $a=3.27m{s}^{-2}$, final velocity $v$ is given by:

$v^2=u^2+2as$

$\Rightarrow v=\sqrt{(10{)}^2+2(3.27\left)\right(10)}\approx 12.86m{s}^{-1}$

$\therefore v\approx 13m/s\to$ option $\text{(a)}$

$\to QED.$