Question

A block of mass $1\text{kg}$ slides down on a rough inclined plane of inclination ${60}^{\circ }$ starting from its top. If the coefficient of kinetic friction is 0.5 and the length of the plane is $1\text{m}$, then work done against friction is (Take $g=9.8{\text{m/s}}^2$)

• A. $9.82\text{J}$
• B. $4.94\text{J}$
• C. $2.45\text{J}$
• D. $1.96\text{J}$

Answer 1

The correct option is C $2.45\text{J}$


From figure, normal reaction is
$N=mg\cos \theta$
$=1\times 9.8\times \cos {60}^{\circ }$
$=4.9\text{N}$
Force due to friction $;f=\mu N$
$=\mu mg\cos {60}^{\circ }$
$=0.5\times 4.9\text{N}$
$=2.45\text{N}$

length of the plane is $S=1\text{m}$.
$\therefore$ Work done against friction$\left(W_f\right)$
$=f\times S=2.45\text{N}\times 1\text{m}$
$=2.45\text{J}$