Question

A spring has natural length 20 cm. Compare the work W1 done in stretching the spring from 20 cm to 30 cm with the work W2 done in stretching it from 30 cm to 40 cm. How are W2 and W1 related?

Answer 1

Answer:

W1 = W2 = 50k Joules (assuming W1 and W2 have the same force constant, k)

W1 - W2 = 0

Step-by-step explanation:

Work done in stretching a spring is given by 1/2ke^2

Where, k is force constant and e is extension

e1 = 30cm - 20cm = 10cm

W1 = 1/2k(e1)^2 = 1/2×k×10^2 = 50k Joules

e2 = 40cm - 30cm = 10cm

W2 = 1/2k(e2)^2 = 1/2×k×10^2 = 50k Joules

W1 = W2 = 50k Joules provided W1 and W2 have equal force constant

Therefore, W1 - W2 = 0