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NEET PHYSICSWORK, ENERGY AND POWEREasy

Question

The minimum work done in pulling up a block of wood weighing 2 kN2 \text{ kN} for a length of 10 m10 \text{ m} on a smooth plane inclined at an angle of 1515^\circ with the horizontal is (given sin15=0.2588\sin 15^\circ = 0.2588):

A

4.36 kJ4.36 \text{ kJ}

B

5.17 kJ5.17 \text{ kJ}

C

8.91 kJ8.91 \text{ kJ}

D

9.82 kJ9.82 \text{ kJ}

Step-by-Step Solution

The minimum force required to pull the block up a smooth inclined plane is equal to the component of its weight along the incline, F=mgsinθF = mg \sin \theta . Given: Weight W=mg=2 kN=2000 NW = mg = 2 \text{ kN} = 2000 \text{ N}, length of incline d=10 md = 10 \text{ m}, and θ=15\theta = 15^\circ. The required force is F=2000×sin15=2000×0.2588=517.6 NF = 2000 \times \sin 15^\circ = 2000 \times 0.2588 = 517.6 \text{ N}. The minimum work done is given by W=F×dW = F \times d . W=517.6 N×10 m=5176 J=5.176 kJ5.17 kJW = 517.6 \text{ N} \times 10 \text{ m} = 5176 \text{ J} = 5.176 \text{ kJ} \approx 5.17 \text{ kJ}. Alternatively, work done equals the change in potential energy =mgh=mg(dsinθ)=2000×10×0.2588=5176 J=5.176 kJ= mgh = mg(d \sin \theta) = 2000 \times 10 \times 0.2588 = 5176 \text{ J} = 5.176 \text{ kJ}.

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 POWERminimumpullingweighinglengthsmooth

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