Question

A block of mass $10 \,kg$, initially at rest, makes a downward motion on $45^{\circ}$ inclined plane. Then the distance travelled by the block after $2 s$ is (Assume the coefficient of kinetic friction to be $0.3$ and $g=10\, ms ^{-2} )$

• A. $7 \sqrt{2}\, m$
• B. $\frac{9}{\sqrt{2}}\, m$
• C. $10 \sqrt{2}\, m$
• D. $5 \sqrt{2} \, m$

Answer 1

Answer: (A)

$7 \sqrt{2}\, m$

Answer 2

The figure given below, shows the downward motion of a body on the inclined plane, then

The equation of acceleration in downwards motion of the block,
$a=g \sin \theta-\mu \,g \cos \theta\,\,\,\,\,\,\dots(i)$
Given, mass of a block, $m=10 \,kg , \theta=45^{\circ}, \mu=0.3$
and time, $t=2\, s$, Putting the given values in Eq (i), we get
$\Rightarrow\, a =g\left(\sin 45^{\circ}-0.3 \cos 45^{\circ}\right)$
$=\frac{10}{\sqrt{2}}(1-0.3)=\frac{10 \times 0.7}{\sqrt{2}} \,m / s ^{2}$
Displacement of the block, from the second equation of the motion,
$s=u t+\frac{1}{2} a t^{2} \,(\because u=0)$
$s=\frac{1}{2} \times \frac{10 \times 0.7}{\sqrt{2}} \times(2)^{2}=\frac{7 \times 4}{2 \sqrt{2}}=7 \sqrt{2} m$