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NEET PHYSICSEasy

Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 m is

A

21 J

B

26 J

C

13 J

D

18 J

Step-by-Step Solution

  1. Identify the Principle: For a variable force, the work done is equal to the area under the Force-Distance (F−dF-d) graph [Class 11 Physics, Ch 6, Sec 5.6, Fig 5.3].
  2. Analyze the Graph (Standard interpretation for this PYQ): The graph typically consists of a rectangular region and a triangular region over the displacement interval.
  • Rectangle: From d=3 md = 3\text{ m} to d=7 md = 7\text{ m} with height F=2 NF = 2\text{ N}. Area1=width×height=(7−3)×2=4×2=8 J\text{Area}_1 = \text{width} \times \text{height} = (7 - 3) \times 2 = 4 \times 2 = 8\text{ J}
  • Triangle: From d=7 md = 7\text{ m} to d=12 md = 12\text{ m} with base (12−7)=5 m(12-7) = 5\text{ m} and height 2 N2\text{ N}. Area2=12×base×height=12×5×2=5 J\text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 2 = 5\text{ J}
  1. Calculate Total Work: W=Area1+Area2=8 J+5 J=13 JW = \text{Area}_1 + \text{Area}_2 = 8\text{ J} + 5\text{ J} = 13\text{ J}
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