Question

answer this question pls ​don't send irrevalant answers otherwise I will report​

Answer 1

The free body diagram of the block is given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){60}}\put(60,0){\line(-4,3){60}}\qbezier(55,0)(54,1.5)(56,3)\put(52,1){ \theta }\multiput(22,28.5)(6,8){2}{\line(4,-3){8}}\multiput(22,28.5)(8,-6){2}{\line(3,4){6}}\put(29,28.5){\vector(0,-1){20}}\put(29,28.5){\vector(4,-3){9.6}}\put(29,28.5){\vector(-3,-4){9.6}}\put(29,28.5){\vector(3,4){9.6}}\put(22,28.5){\vector(-4,3){8}}\put(26,4.5){\sf{mg}}\put(6,11){ \sf{mg\cos\theta} }\put(39.5,41){ \sf{R} }\put(39.5,19){ \sf{mg\sin\theta} }\put(16,35){\sf{f}}\end{picture}$

The frictional force acting on the block is,

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

From the free body diagram of the block,

$\longrightarrow\sf{f=\mu\,mg\cos\theta}$

$\longrightarrow\sf{f=0.75\,mg\cos\theta}$

$\longrightarrow\sf{f=\dfrac{3}{4}\,mg\cos\theta}$

Since $\sf{\dfrac{3}{4}\,mg}$ is a constant,

$\longrightarrow\sf{f\propto\cos\theta}$

For $\sf{\theta=0^o,}$

$\longrightarrow\sf{f=\dfrac{3}{4}\,mg\cos0^o}$

$\longrightarrow\sf{f=\dfrac{3}{4}\,mg}$

This indicates $\sf{f\neq0}$ for $\sf{\theta=0^o}$ unless $\sf{m=0\ kg.}$

Hence the 2nd graph will be correct as $\sf{f=0}$ for $\sf{\theta=0^o}$ in other graphs.

Since $\cos\theta\propto\dfrac{1}{\theta}$ for $\sf{0^o\leq\theta\leq37^o,}$ the graph should be strictly decreasing in the interval $\sf{[0^o,\ 37^o],}$ which is possible only in the 2nd graph.

Hence 2nd graph is the answer.