Question

A block of mass $5\text{kg}$ slides down on a rough inclined plane of coefficient of friction $0.75$ having angle of inclination ${37}^{\circ }$ as shown in the figure. Initially if the block is at rest on the top of inclined plane of length $2\text{m}$, then the work done against the friction in moving the block to the bottom of inclined plane is (Take $g=10{\text{m/s}}^2$).

• A. $0\text{J}$
• B. $5\text{J}$
• C. $10\text{J}$
• D. $-5\text{J}$

Answer 1

The correct option is A $0\text{J}$

Given, mass of the block $\left(m\right)=5\text{kg}$
Length of the inclined plane $\left(s\right)=2\text{m}$
By the FBD of the block:
$R=mg\cos {37}^{\circ }$
$(f_s{)}_{\text{max}}={\mu }_{s}R$
$=0.75\times mg\cos {37}^{\circ }$
$=\frac{3}{4}\times 5\times 10\times \frac{4}{5}$
$(f_s{)}_{\text{max}}=30\text{N}$ and the applied force is $mg\sin {37}^{\text{o}}=5\times 10\times \frac{3}{5}=30\text{N}$

Here force due to friction is sufficient to hold the block at rest
$\therefore$ Block does not slide down.
Hence, displacement $=0$
$\Rightarrow$ Workdone against friction $=0$

Hence option A is the correct answer