Question

A block of mass 500 g slides down on a rough incline plane of inclination 53° with a uniform speed. Find the work done against the friction as the block slides through 2 m. [g = 10 m/s²]

Answer 1

8 J

Answer 2

The block slides down the incline at a constant speed, meaning the net force along the incline is zero. The forces acting along the incline are the component of gravity down the incline ($mg \sin \theta$) and the kinetic friction force ($f_k$) up the incline. For zero net force, these two forces must be equal in magnitude: $f_k = mg \sin \theta$. Using $m=0.5$ kg, $g=10$ m/s², and $\sin 53^\circ \approx 0.8$, we find $f_k = (0.5)(10)(0.8) = 4$ N. The block slides through a distance of $d=2$ m. The friction force acts opposite to the displacement. The work done by friction is $W_{friction} = f_k \times d \times \cos(180^\circ) = 4 \times 2 \times (-1) = -8$ J. The work done against friction is the negative of the work done by friction, which is $-(-8 \text{ J}) = 8$ J.