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 to 40 cm. (Use k for the spring constant) How are W2 and W1 related?

W₂ is three times W₁ (W₂ = 3W₁)

Step-by-step explanation:

Applying,

W = ke²/2 Equation 1

Where W = workdone in stretching the spring, k = spring constant, e = extension.

For W₁,

W₁ = ke₁²/2

Given: e₁ = 30-20 = 10 cm = 0.1 m

Substitute these value into equation 1

W₁ = k(0.1²)/2

W₁ = 0.005k Joules

For W₂,

W₂ = (ke/2)-W₁

Given: e = (40-20) = 20 cm = 0.1 m

Substitute these value into equation 1

W₂ = (k×0.2²/2)-0.005

W₂ = 0.015k Joules.

W₂/W₁ = 0.015k/0.005k

W₂/W₁ = 3

Therefore,

W₂ = 3W₁

450 is the final answer! : )

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step-by-step explanation: