Q: A body of mass ‘m’ slides down a smooth inclined plane having an inclination of 45° with the horizontal. It takes 2s to reach the bottom. If the body is placed on a similar plane having coefficient of friction 0.5 then what is the time taken for it to reach the bottom ?
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Sol: Mass = m , θ = 45°, μ = 0.5
Time taken by the body to reach the bottom without friction is
$\large t_1 = \sqrt{\frac{2l}{g sin\theta}} $ …(i)
Time taken in friction is
$\large t_2 = \sqrt{\frac{2l}{g (sin\theta – \mu cos\theta }} $ …(ii)
On dividing ,
$\large \frac{t_2}{t_1} = \sqrt{\frac{sin\theta}{sin\theta – \mu cos\theta}}$
$\large t_2 = t_1 \sqrt{\frac{sin\theta}{sin\theta – \mu cos\theta}}$
$\large t_2 = 2 \sqrt{\frac{sin45^o}{sin45^o – 0.5 \times cos45^o}}$
= 2√2 sec