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

Answer 1

The work done to stretch the spring from 30 to 40 cm (W2) is four times the work done to stretch it from 20 to 30 cm (W1), because the work done on a spring is proportional to the square of the displacement from its natural length.

Step-by-step explanation:

To compare the work W1 done in stretching the spring from its natural length of 20 cm to 30 cm with the work W2 done in stretching it from 30 cm to 40 cm, we can use the formula for the work done on a spring: W = 1/2kx², where k is the spring constant, and x is the displacement from the natural length. The displacement from 20 cm to 30 cm is 10 cm (0.1 m), and from 30 cm to 40 cm is also 10 cm (0.1 m). Using the variables: x1 = 0.1 m for the first stretch, and x2 = 0.2 m for the second stretch (since it's from the natural length).

Calculating the work for each stretch:

• W1 = 1/2k(x1)² = 1/2k(0.1²)
• W2 = 1/2k(x2)² = 1/2k(0.2²)

Since x2 is double the first displacement (x1), the square of x2 (0.2²) is four times the square of x1 (0.1²). So the work done to stretch the spring from 30 to 40 cm (W2) is four times the work done to stretch it from 20 to 30 cm (W1). Thus, W2 = 4W1.

Answer 2

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