A block of mass $m$ is placed on a rough plane with coefficient of friction $\mu$, whose angle of in

Question

A block of mass $m$ is placed on a rough plane with coefficient of friction $\mu$, whose angle of inclination with the horizontal can be varied as $0 \leq 	heta<\pi / 2$. If the block is tied with an elastic spring of spring constant $k$ and $x$ be the instantaneous deformation in the spring, then the correct variation of $x$ versus $	heta$ is given by

Professor PC Thomas Classes · Answer

If $	heta \leq 	an ^{-1} \mu$, block is not sliding downward, therefore, the spring is not stretched, thus $x=0$. If $	heta>	an ^{-1} \mu$, the spring gets stretched by $ x=\frac{m g}{k}(\sin 	heta-\mu \cos 	heta) $. \- First Derivative (Slope): $\frac{d x}{d 	heta}=\frac{m g}{k}(\cos 	heta+\mu \sin 	heta)$. Since $	heta$ is between 0 and $\pi / 2$, the slope is positive (the graph goes up).
- Second Derivative (Concavity): $\frac{d^2 x}{d 	heta^2}=\frac{m g}{k}(-\sin 	heta+\mu \cos 	heta)$.
Since we are looking at the region where $	an 	heta>\mu$, it follows that $\sin 	heta>\mu \cos 	heta$. Therefore:
$$
\frac{d^2 x}{d 	heta^2}<0
$$A negative second derivative indicates that the graph is concave down (it "bends" toward the $	heta$-axis).