The net horizontal force acting on a cart varies with displacement as follows

PHYSICS · Multi-part · CBSE · Physics · Work, Energy and Power · Work, Energy and Power

Problem

The net horizontal force acting on a cart varies with displacement as follows. Starting from x=0x = 0, the force rises linearly from 0 N0\,\text{N} at x=0x = 0 to 12 N12\,\text{N} at x=2 mx = 2\,\text{m}, remains constant at 12 N12\,\text{N} up to x=4 mx = 4\,\text{m}, then falls linearly to −6 N-6\,\text{N} at x=6 mx = 6\,\text{m}, and finally returns linearly to 0 N0\,\text{N} at x=7 mx = 7\,\text{m}. The cart starts from rest at x=0x = 0. (i) Reproduce a neat labelled sketch of the force–displacement graph described above and divide the area under the curve into simple geometric figures. (ii) Calculate the work done in each region identified in part (i) and hence find the net work performed on the cart over the entire 7 m7\,\text{m} displacement. (iii) If the same net work were to be done by a single constant horizontal force acting over the 7 m7\,\text{m} distance, compute the magnitude and sign of that constant force. (iv) Without performing any further calculations, state whether the kinetic energy of the cart at x=7 mx = 7\,\text{m} is greater than, less than, or equal to its initial kinetic energy, and justify your answer.

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