Physics Circle home Student Login

Home›Chapters›Work, Energy and Power PYQs›Spring & Elastic Energy›Question 32

JEE Main 2003 (AIEEE) · Spring & Elastic Energy

Spring & Elastic Energy question from JEE Main 2003 (AIEEE)

Q32JEE Main 2003 · AIEEEEasy

Spring & Elastic Energy

A spring of spring constant 5×103 N/m5 \times 10^{3}\text{ N/m} is stretched initially by 5 cm5\text{ cm} from the unstretched position. Then the work required to stretch it further by another 5 cm5\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=0.10x_2 = 0.10 m.
  2. W=12(5000)(0.01−0.0025)=2500×0.0075W = \frac{1}{2}(5000)(0.01 - 0.0025) = 2500 \times 0.0075
  3. W=18.75W = 18.75 N m
  4. So the answer is (b) 18.7518.75 N-m.
Report a mistake

More questions on Spring & Elastic Energy