A block of mass 2 kg slides down an incline p | vedclass

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Question

Given,

$m =2\,kg$

$	heta=30^{\circ}$

$\mu=0.5$

$g =10\,m / s ^2$

The contact force is $F _{ i }$ is the resultant of frictional force

Vedclass · Accepted Answer

$20 \,Nt$

Vedclass · Answer

Given,

$m =2\,kg$

$	heta=30^{\circ}$

$\mu=0.5$

$g =10\,m / s ^2$

The contact force is $F _{ i }$ is the resultant of frictional force and normal force.

From the free body diagram,

Normal force, $N = mg \cos 	heta$

$N=2 	imes 10 	imes \cos 30^{\circ}$

$N=17.320\,N$

The friction force, $F _{ r }=\mu \mu g \cos 	heta$

$F_r=0.5 	imes 2 	imes 	imes 10 	imes \cos 30^{\circ}$

$F _{ r }=8.660\,N$

$N$ and $F_r$ are perpendicular to each other.

The contact force is, $F_c=\sqrt{N^2+F_r^2}$

$F_c=\sqrt{(17.320)^2+(8.699)^2}$

$F _{ c }=19.381\,N$

We can say approximately $20\,N$

A block of mass $2 kg$ slides down an incline plane of inclination $30^o$. The coefficient of friction between block and plane is $0.5$. The contact force between block and plank is :

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