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NEET PHYSICSWORK, ENERGY AND POWEREasy

Question

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g = 10 m/s², work done against friction is:

A

200 J

B

100 J

C

zero

D

1000 J

Step-by-Step Solution

To find the work done against friction, we use the principle of conservation of energy. The total work done (WtotalW_{total}) by an external force to move the block up the incline is equal to the increase in the block's gravitational potential energy (ΔU\Delta U) plus the work done against dissipative forces like friction (WfrictionW_{friction}).

  1. Calculate the increase in Potential Energy (\Delta U): The potential energy gained by raising a mass mm to a height hh is given by the formula U=mghU = mgh , . ΔU=2 kg×10 m/s2×10 m=200 J\Delta U = 2 \text{ kg} \times 10 \text{ m/s}^2 \times 10 \text{ m} = 200 \text{ J}

  2. Calculate work done against friction (WfrictionW_{friction}): Using the energy balance equation: Wtotal=ΔU+WfrictionW_{total} = \Delta U + W_{friction} 300 J=200 J+Wfriction300 \text{ J} = 200 \text{ J} + W_{friction} Wfriction=300 J200 J=100 JW_{friction} = 300 \text{ J} - 200 \text{ J} = 100 \text{ J}

Thus, 100 J of the total work was used to overcome friction.

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.

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