A) 10 N
B) 15 N
C) 16.3 N
D) 25 N
Correct Answer: C
Solution :
\[a=\frac{g\sin \theta }{1+\frac{{{k}^{2}}}{{{r}^{2}}}}=\frac{g\sin \theta }{1+\frac{1}{2}}=\frac{2}{3}g\sin \theta \] Minimum frictional force \[=F=mg\sin \theta -ma\] \[=mg\sin \theta -m\frac{2}{3}g\sin \theta \] \[=\frac{1}{3}mg\sin \theta \] \[F=\frac{1}{3}\times 10\times 10\times \frac{1}{2}\] \[=\frac{50}{3}=16.3\,N\]
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