An object of mass 2 kg , initially at rest, moves along the x-axis under the action of a variable force. The force-position ( F-x ) graph consists of a triangular region from x = 0 to x = 4 m where the force increases linearly from zero to a maximum value F₀ at x = 2 m and then decreases linearly to zero at x = 4 m . This is followed by a rectangular region from x = 4 m to x = 6 m where the force is constant at - F₀

Step-by-step solution

According to the work-energy theorem, the net work done by the force equals the change in kinetic energy of the object. The work done is the area under the F-x graph. Area of the triangular region (from x = 0 to x = 4 m ) = 1 2 base height = 1 2 4 F₀ = 2F₀ Area of the rectangular region (from x = 4 m to x = 6 m ) = width height = 2 (- F₀ 2 ) = -F₀ Net work done W = 2F₀ - F₀ = F₀ Change in kinetic energy K = 1 2 mv^2 - 0 = 1 2 2 (4)^2 = 16 J Equating work done to the change in kinetic energy: F₀ = 16 N Answer: 16 N