A block 'A' takes 2,s to slide down a frictio | vedclass

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Question

(C) Since the lift is moving with uniform velocity,its acceleration is zero. Therefore,the effective acceleration due to gravity on the block remains

Vedclass · Accepted Answer

$1.68$

Vedclass · Answer

(C) Since the lift is moving with uniform velocity,its acceleration is zero. Therefore,the effective acceleration due to gravity on the block remains '$g$'.For a block sliding down a frictionless incline,the acceleration is $a = g \sin 	heta$.Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ and $s = l$:$l = \frac{1}{2} (g \sin 30^{\circ}) (2)^2 = \frac{1}{2} g (0.5) (4) = g$.Now,for the incline of $45^{\circ}$,let the time taken be '$t$':$l = \frac{1}{2} (g \sin 45^{\circ}) t^2$.Since '$l$' is constant,we equate the two expressions:$g = \frac{1}{2} g \left(\frac{1}{\sqrt{2}}ight) t^2$.$1 = \frac{1}{2\sqrt{2}} t^2 \Rightarrow t^2 = 2\sqrt{2} \approx 2.828$.$t = \sqrt{2.828} \approx 1.68\,s$.

$A$ 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 $........\,s$.

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