Question

See the attachment for question 30 please answer fast with solution and answer is A

Answer 1

$\setlength{\unitlength}{0.9mm}\begin{picture}(5,5)\put(0,0){\framebox(15,15){\sf{2.5\ kg}}}\put(15,7.5){\vector(1,0){15}}\put(31.5,6){\sf{6\ N}}\put(7.5,15){\vector(0,1){15}}\put(2.6,31.5){ \sf{P+R} }\put(7.5,0){\vector(0,-1){15}}\put(3,-19){\sf{25\ N}}\put(0,0.2){\vector(-1,0){20}}\put(-22,1){\sf{f}}\put(-30,0){\line(1,0){75}}\multiput(-30,0)(3,0){26}{\qbezier(0,0)(-1,-1)(-2,-2)}\end{picture}$

The net vertical force acting on the block should be zero. Thus,

$\longrightarrow\sf{P+R-25=0}$

$\longrightarrow\sf{R=25-P}$

If $\sf{P=9\ N,}$

$\longrightarrow\sf{R=25-9}$

$\longrightarrow\sf{R=16\ N}$

The coefficient of static friction for the block and surface is,

• $\sf{\mu=0.4}$

Hence the friction force acting on the block is,

$\longrightarrow\sf{f=\mu\,R}$

$\longrightarrow\sf{f=0.4\times16\ N}$

$\longrightarrow\underline{\underline{\sf{f=6.4\ N}}}$

Hence (B) is the answer.