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stiv31 [10]
3 years ago
5

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

n stretching it from 30 to 40 cm. (Use k for the spring constant) How are W2 and W1 related?
Mathematics
1 answer:
Ksju [112]3 years ago
3 0

Answer:

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

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