Question

A block of mass $0.5\, kg$ has an initial velocity of $10\, m\, s ^{-1}$ down an inclined plane of angle $30^{\circ},$ the coefficient of friction between the block and the inclined surface is $0.2$ The velocity of the block after it travels a distance of $10 m$ is (Take $\left.g=10\, m\, s \,{}^{-2}\right)$

• A. $17 \,m\, s ^{-1}$
• B. $13\, m\, s ^{-1}$
• C. $24\, m\, s ^{-1}$
• D. $8\, m\, s ^{-1}$

Answer 1

Answer: (D)

Here, $m=0.5\, kg , u=10\, ms ^{-1}, \theta=30^{\circ}, \mu=0.2, s=10 m$
Acceleration down the plane,
$ a=g(\sin \theta-\mu \cos \theta)=10\left(\sin 30^{\circ}-0.2 \cos 30^{\circ}\right)=3.268 $
$\text { From, } v^{2}=u^{2}+2 a s=10^{2}+2(3.268) \times 10=165.36 $
or $\quad v=\sqrt{165.36} \approx 13\, m\, s ^{-1}$