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 i

Question

3 years ago

6

n stretching it from 30 cm to 40 cm. How are W2 and W1 related?

1 answer:

sashaice [31] · Answer 1 · 2022-03-05T02:39:32+03:00

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