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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

Force FF on a particle moving in a straight line varies with distance dd as shown in the figure. The work done on the particle during its displacement of 1212 m is:

A

21 J

B

26 J

C

13 J

D

18 J

Step-by-Step Solution

According to the principles of physics, the work done by a variable force is equivalent to the area under the Force-displacement (FdF-d) curve . Based on the standard graphical data for this specific problem (AIPMT 2011), the total work is calculated by summing the areas of the individual sections:

  1. Area from d=0d = 0 to 33 m: This is a rectangle with height 22 N and width 33 m. Area =3×2=6= 3 \times 2 = 6 J.
  2. Area from d=3d = 3 to 77 m: This is a rectangle with height 33 N and width 44 m (73=47 - 3 = 4). Area =4×3=12= 4 \times 3 = 12 J.
  3. Area from d=7d = 7 to 1212 m: This section lies below the distance axis, indicating a negative force of 1-1 N over a width of 55 m (127=512 - 7 = 5). Area =5×(1)=5= 5 \times (-1) = -5 J.

Total Work Done (WW) =6 J+12 J5 J=13= 6 \text{ J} + 12 \text{ J} - 5 \text{ J} = 13 J.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWORK, ENERGY AND POWERparticlemovingstraightvariesdistance

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