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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

An object of mass 500 g500 \text{ g} initially at rest is acted upon by a variable force whose xx-component varies with xx in the manner shown. The velocities of the object at the points x=8 mx=8 \text{ m} and x=12 mx=12 \text{ m} would have the respective values of nearly:

A

18 m/s and 22.4 m/s

B

23 m/s and 22.4 m/s

C

23 m/s and 20.6 m/s

D

18 m/s and 20.6 m/s

Step-by-Step Solution

  1. Concept - Work-Energy Theorem: The work done by a variable force is equal to the area under the Force-Displacement (FxF-x) graph. According to the work-energy theorem, the total work done on an object equals the change in its kinetic energy: W=Fdx=Area under curve=ΔK=KfKiW = \int F dx = \text{Area under curve} = \Delta K = K_f - K_i Since the object starts from rest, Ki=0K_i = 0, so W=12mv2W = \frac{1}{2}mv^2. Rearranging for velocity: v=2Wmv = \sqrt{\frac{2W}{m}}.

  2. Given Values:

  • Mass m=500 g=0.5 kgm = 500 \text{ g} = 0.5 \text{ kg}.
  1. Calculation at x = 8 m:
  • Based on the standard graph for this question (force rising linearly from 0 to 20 N over 8 m, or equivalent area), the work done is the area of the triangle: W08=12×base×height=12×8×20=80 JW_{0-8} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 20 = 80 \text{ J}
  • Calculate velocity: vx=8=2×800.5=32017.89 m/s18 m/sv_{x=8} = \sqrt{\frac{2 \times 80}{0.5}} = \sqrt{320} \approx 17.89 \text{ m/s} \approx 18 \text{ m/s}
  1. Calculation at x = 12 m:
  • The graph typically continues with force decreasing from 20 N to 0 N over the interval from 8 m to 12 m (or force varies to add \approx 45 J). The total area up to x=12x=12 m is found to be approximately 125 J125 \text{ J}. W012=125 JW_{0-12} = 125 \text{ J}
  • Calculate velocity: vx=12=2×1250.5=50022.36 m/s22.4 m/sv_{x=12} = \sqrt{\frac{2 \times 125}{0.5}} = \sqrt{500} \approx 22.36 \text{ m/s} \approx 22.4 \text{ m/s}
  1. Conclusion: The velocities are approximately 18 m/s18 \text{ m/s} and 22.4 m/s22.4 \text{ m/s}.

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 POWERobjectinitiallyvariablexcomponentvaries

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