Question

A block of mass $15\text{kg}$ is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of $50\text{N}$. The coefficient of friction between the surfaces of contacts is ($g=10{\text{m/s}}^2$)

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{1}{2}$


$T=50\text{N}$
Component of force (in y direction)
$N=\frac{T}{\sqrt{2}}+\frac{150}{\sqrt{2}}\Rightarrow N=200/\sqrt{2}$
$f=\mu \frac{200}{\sqrt{2}}$ (As we know, f = $\mu$N)
Component of force in x direction,
$\frac{150}{\sqrt{2}}=\frac{T}{\sqrt{2}}+f,$
$\frac{150}{\sqrt{2}}=\frac{50}{\sqrt{2}}+\frac{\mu \times 200}{\sqrt{2}}\Rightarrow \mu =\frac{1}{2}$

Why this question? Tip: For questions based on motion along inclined plane, it is better to define the axises along the incline and perpendicular to the incline.