A body starts from rest on a long inclined pl | vedclass

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(D) The body moves down an inclined plane with an angle of inclination $	heta = 45^o$.The forces acting on the body along the plane are the component

Vedclass · Accepted Answer

$3.33$

Vedclass · Answer

(D) The body moves down an inclined plane with an angle of inclination $	heta = 45^o$.The forces acting on the body along the plane are the component of gravity $mg \sin 	heta$ acting downwards and the frictional force $f = \mu N$ acting upwards.The normal force is $N = mg \cos 	heta$.The net acceleration $a$ of the body is given by $a = g \sin 	heta - \mu g \cos 	heta$.For the speed to be maximum,the acceleration must be zero $(a = 0)$.Setting $a = 0$,we get $g \sin 	heta = \mu g \cos 	heta$,which simplifies to $\mu = 	an 	heta$.Given $\mu = 0.3x$ and $	heta = 45^o$,we have $0.3x = 	an 45^o$.Since $	an 45^o = 1$,we get $0.3x = 1$.Therefore,$x = \frac{1}{0.3} = 3.33 \ m$.

$A$ body starts from rest on a long inclined plane of slope $45^o$. The coefficient of friction between the body and the plane varies as $\mu = 0.3x$,where $x$ is the distance travelled down the plane. The body will have maximum speed (for $g = 10 \ m/s^2$) when $x = $ ........ $m$.

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