A spring of force constant k is stretched by a length x. The work done in stretching it further by the same length x is: MCQ with Answer and Explanation
A spring of force constant k is stretched by a length x. The work done in stretching it further by the same length x is:
A. ³/₂kx²
B. kx²
C. 2kx²
D. ½kx²
Answer: Option A
Solution (By JKSSB Mock Tests)
Work done to stretch spring from 0 to x: W₁ = ½kx². Work done from 0 to 2x: W₂ = ½k(2x)² = 2kx². Thus work for additional stretch from x to 2x: ΔW = W₂ - W₁ = 2kx² - ½kx² = ³/₂kx². Spring force is variable (F=kx), so work is integral of F·dx, yielding parabolic energy storage. Memory aid: Elastic potential energy U = ½kx²; always calculate difference for incremental work. This problem tests understanding of work done by variable forces, a common theme in energy conservation questions in competitive exams.
Explanation:
Total height = 30 × 0.2 m = 6 m. Work done = mgh = 50 × 10 × 6 = 3000 J. Power = Work/time = 3000/10 = 300 W. If steps height and number given, calculate total vertical displacement. Time taken for climbing is important. Power is rate of doing work.
Explanation:
Myopic eye focuses image in front of retina. A diverging (concave) lens spreads light so image falls on retina. Hypermetropia (farsightedness) corrected by convex lens. Bifocals for presbyopia. Cylindrical for astigmatism.
Explanation:
Acceleration phase: s₁ = ½a t² = ½×2×100 = 100 m, v = a t = 20 m/s. Constant velocity phase: s₂ = v×t = 20×10 = 200 m. Total = 300 m. Motion in two parts.
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