A body of mass 5 kg moves along a straight line under the action of a variable force. From x = 0 to x = 4 m, the force acts in the direction of displacement and increases linearly from 0 N to 20 N. From x = 4 m to x = 6 m, the motion of the body reverses while the force acts in forward direction. From x = 6 m to x = 10 m, the force becomes negative acting opposite to the original direction of motion. Calculate the net work done by the force on the body.

Work done by a variable force equals the area under the force–displacement (F–x) graph and its sign depends on the direction of displacement relative to the force.

From x = 0 m to x = 4 m:

The force increases linearly from 0 N to 20 N in the direction of displacement so, the graph forms a triangle and the work done is given by,

Area of the triangle formed = Work done = $\dfrac{1}{2}$ × 4 × 20
= +40 J [∵ Force acts along the direction of displacement]

From x = 4 m to x = 6 m:

The motion of the body reverses while the force still acts forward. Since displacement is opposite to the force, the work done is negative. Assuming the force remains 20 N, the area formed is a rectangle and the work done is given by,

Area of the rectangle formed = Work done = -20 × 2 = -40 J,

From x = 6 m to x = 10 m

The force becomes negative (opposite to the original direction) and since the displacement is also reversed so both force and displacement are in the same direction and the work done again becomes positive. The graph forms a rectangle and the work done is given by.

Area of the rectangle formed = Work done = 4 × 20 = 80 J

Net work done by the force = 40 + (-40) + 80 = 120 - 40 = 80 J

Hence, the net work done by the force on the body is 80 J.