Q 11-05-086JEE MainJEE Main 2024 (30 Jan, Shift 2)Easy
A block of mass $1\ \text{kg}$ is pushed up a surface inclined to the horizontal at an angle of $60^\circ$ by a force of $10\ \text{N}$ parallel to the inclined surface. The coefficient of friction between the block and the surface is $0.1$. When the block is pushed up by $10\ \text{m}$ along the inclined surface, the work done against the frictional force is: ($g = 10\ \text{m s}^{-2}$)
Answer: (B) $5\ \text{J}$
$f = \mu mg\cos60^\circ = 0.1\times1\times10\times\tfrac12 = 0.5\ \text{N}$
$$W = f\,s = 0.5\times10 = 5\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics