Physics Circle home Student Login

Home›Chapters›Work, Energy and Power PYQs›Spring & Elastic Energy›Question 33

JEE Main 2002 (AIEEE) · Spring & Elastic Energy

Spring & Elastic Energy question from JEE Main 2002 (AIEEE)

Q33JEE Main 2002 · AIEEEEasy

Spring & Elastic Energy

A spring of force constant 800 N/m800\text{ N/m} has an extension of 5 cm.5\text{ cm}. The work done in extending it from 5 cm5\text{ cm} to 15 cm15\text{ cm} is

Show answer and solution

Answer: (b)

  1. W=12k(x22−x12)W = \frac{1}{2}k(x_2^2 - x_1^2) with x1=0.05x_1 = 0.05 m, x2=0.15x_2 = 0.15 m.
  2. W=12(800)(0.0225−0.0025)=400×0.02W = \frac{1}{2}(800)(0.0225 - 0.0025) = 400 \times 0.02
  3. So the answer is (b) 88 J.
Report a mistake

More questions on Spring & Elastic Energy