Question

A 30 kg box has to move up an inclined slope of $ 30{}^\circ $ to the horizontal at a uniform velocity of 5 m/s. If the frictional force retarding the motion is 250 N the horizontal force in newtons to move up is $ (g=10\text{ }m/{{s}^{2}}) $ :

• A. 400 $ \sqrt{2} $ N
• B. 400 N
• C. $ \frac{400\sqrt{3}}{2}N $
• D. $ 400\times 2\sqrt{3}N $

Answer 1

Answer: (C)

A box is moving with a uniform velocity, then the force acting on the body is given by $ F=mg\sin \theta +\mu mg\cos \theta $ $ =30\times 10\times \sin {{30}^{o}}+250 $ $ =30\times 10\times \frac{1}{2}+250=400 $ Now, horizontal component of force is given by $ F=\cos \theta =400\times \cos {{30}^{0}}=\frac{400\sqrt{3}}{2}N $