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Grade: 11
        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

Answers : (1)

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}
similarly
t\sqrt{\frac{2s}{a}}
t_{2} = \sqrt{\frac{2s}{\frac{g}{\sqrt{2}}}}
as per question
t1 = 2t2
\mu =\frac{0.75}{\sqrt{2}}
2 years ago
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