Question

A 30 kg box has to move up an inclined slope of ${30}^0$ to horizontal at a uniform velocity of 5 m/sec. If ${\mu }_k=\frac{5}{3\sqrt{3}}$, the horizontal force required to move up is $(g=10m/se{c}^2)$

• A. $300\sqrt{2}N$
• B. $300N$
• C. $\frac{300\sqrt{3}}{2}N$
• D. $300\times 2\sqrt{3}N$

Answer 1

Answer: (D)

$300\times 2\sqrt{3}N$

Let the horizontal force to move up be F.
F will have a component
$F\cos {30}^o$ up the incline and a component $F\sin {30}^o$ down to balance the forces along the inclined plane, $Fcos{30}^o=mgsin{30}^o+\mu (Fsin{30}^o+mgcos{30}^o)$
$\therefore F(\cos {30}^0-\mu \sin {30}^0)=mg(\sin {30}^0+\mu \cos {30}^0)$
$\Rightarrow F\times (\frac{\sqrt{3}}{2}-\frac{5}{3\sqrt{3}}\times 0.5)=30\times 10(\frac{1}{2}+\frac{5}{3\sqrt{3}}\times \frac{\sqrt{3}}{2})$

Simplifying, $F=300\times 2\sqrt{3}N$