A) 400 \[\sqrt{2}\]N
B) 400 N
C) \[\frac{400\sqrt{3}}{2}N\]
D) \[400\times 2\sqrt{3}N\]
Correct Answer: C
Solution :
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 \[F=\cos \theta =400\times \cos {{30}^{0}}=\frac{400\sqrt{3}}{2}N\]
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