Question

A mass of $0.5\, kg$ is just able to slide down the slope of an inclined rough surface when the angle of inclination is $60^{\circ}$. The minimum force necessary to pull the mass up the incline along the line of greatest slope is : (Take $g=10 \, m / s ^{2}$ )

• A. $20\, N$
• B. $9\, N$
• C. $100 \, N$
• D. $1 \, N$

Answer 1

Answer: (B)

$\mu m g \cos 60^{\circ}=m g \sin 60^{\circ}$
$\mu=\sqrt{3}$
Now, if force $F$ is applied to pull it along the incline,
$F=m g \sin 60^{\circ}+\mu m g \cos 60^{\circ} $
$F=\frac{5 \sqrt{3}}{2}+\frac{5 \sqrt{3}}{2} $
$F=5 \sqrt{3} N =8.66\, N \approx 9\, N$