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        A body slides down an inclined with 45 inclination, twice the time taken to slide is equal to the time taken for the same distance in absence of friction.find co efficient of friction
2 years ago

Rahul Verma
45 Points
							In presence of friction,$mg\sin 45 - \mu N = ma$$= mg/\sqrt{2} - \mu mg = ma$$a = g/\sqrt{2} - \mu g$time taken to slide ($t_{1}$), using eq of motion,$s = \frac{1}{2} at_{1}^{2}$$t_{1} = \sqrt{\frac{2s}{a}}$$t_{1} = \sqrt{\frac{2s}{\frac{g}{\sqrt{2} }- \mu g}}$ in abscence of friction $mg\sin 45 = ma$$a = g\sin 45$$a = g/\sqrt{2}$similarlyt2 = $\sqrt{\frac{2s}{a}}$$t_{2} = \sqrt{\frac{2s}{\frac{g}{\sqrt{2}}}}$as per questiont1 = 2t2$\mu =\frac{0.75}{\sqrt{2}}$

2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions