A block 'A' takes 2 s to slide down a frictionless incline of 30$^\circ$ and length 'l', kept inside a lift going up with uniform velocity 'v'. If the incline is changed to 45$^\circ$, the time taken by the block, to slide down the incline, will be approximately :
Solution
<p>${\theta _1} = 30^\circ ,\,{\theta _2} = 45^\circ$</p>
<p>${a_1} = g\sin {\theta _1} = 5$ m/s<sup>2</sup>, ${a_2} = g\sin {\theta _2} = 5\sqrt 2$ m/s<sup>2</sup></p>
<p>$${{{t_1}} \over {{t_2}}} = {{\sqrt {{{2l} \over {{a_1}}}} } \over {\sqrt {{{2l} \over {{a_2}}}} }} = \sqrt {{{{a_2}} \over {{a_1}}}} $$</p>
<p>${{{t_1}} \over {{t_2}}} = {(2)^{1/4}}$</p>
<p>${t_2} = {(2)^{3/4}}$</p>
<p>$\approx 1.68$ s</p>
About this question
Subject: Physics · Chapter: Laws of Motion · Topic: Newton's First Law
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