$A$ boy is moving on a straight road against a frictional force of $5\, N$. After travelling a distance of $1.5\, km$ he reaches a roundabout (Fig.) of radius $100\, m$. He moves on the circular path for one and a half cycles and then moves forward up to $2.0\, km$. Calculate the total work done by him against the frictional force.

(N/A) The work done by a force is given by the product of the force and the displacement in the direction of the force.
Work done $W = F \times d$
Here,the boy is moving against a constant frictional force $F = 5\, N$ throughout the entire path.
Total distance travelled by the boy is the sum of the straight path,the circular path,and the final straight path.
$1$. Initial straight path distance $= 1.5\, km = 1500\, m$.
$2$. Circular path distance: The boy completes $1.5$ cycles on a circular path of radius $r = 100\, m$. Distance $= 1.5 \times (2 \pi r) = 1.5 \times 2 \times 3.14 \times 100 = 942\, m$.
$3$. Final straight path distance $= 2.0\, km = 2000\, m$.
Total distance $d = 1500\, m + 942\, m + 2000\, m = 4442\, m$.
Since the frictional force acts along the entire path length,the work done against friction is calculated using the total distance covered.
Work done $W = 5\, N \times 4442\, m = 22210\, J$.