Question

A $30kg$ box has to move up an inclined slope of ${30}^o$ to the horizontal at a uniform velocity of $5m{s}^{-1}$. If the frictional force retarding the motion is $150N$, the horizontal force required to move up is $(g=10m{s}^{-2})$

• A. $300\times \frac{2}{\sqrt{3}}N$
• B. $300\times \frac{\sqrt{3}}{2}N$
• C. $300N$
• D. $150N$

Answer 1

Answer: (C)

$300N$

Given: a 30kg box has to move up an inclined slope of ${30}^o$ to the horizontal at a uniform velocity of $5m/s$. If the frictional force retarding the motion is 150N

To find the horizontal force required to move up

Solution:

From the above figure,

Force of friction = 150N

Net horizontal force,

$F-f=mg\sin \theta ⟹F=mg\sin {30}^{\circ }+f⟹30\times 10\times \frac{1}{2}+150⟹F=300N$

is the horizontal force required to move up