Starting from rest a body slides down a 45^{c | vedclass

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Question

Let length is $\ell$ of inclined plane, then $f_r=\mu N =\mu mg \cos 	heta$

$mg \sin 	heta- f _{ r }= ma$

$mg \sin 	heta-\mu mg \cos 	heta=

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

Let length is $\ell$ of inclined plane, then $f_r=\mu N =\mu mg \cos 	heta$

$mg \sin 	heta- f _{ r }= ma$

$mg \sin 	heta-\mu mg \cos 	heta= ma$

Now

$\ell=\frac{1}{2} at ^2=\frac{1}{2} g (\sin 	heta-\mu \cos 	heta) t ^2$

$	ext { Now } \ell_1=\ell_2$

$\ell_2=\frac{1}{2} g \sin 	heta\left(\frac{ t ^2}{2}ight)$

$\frac{1}{2} g(\sin 	heta-\mu \cos 	heta) t ^2=\frac{1}{2} g(\sin 	heta-0 	imes \cos 	heta)\left(\frac{ t ^2}{2}ight)$

$4(\sin 	heta-\mu \cos 	heta)=\sin 	heta$

$\mu=\frac{3}{4}=0.75$

Starting from rest a body slides down a $45^{\circ}$ ind ined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:

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