Question

A spring with spring constant of 33 N/m is stretched 0.15 m from its equilibrium position. How much work must be done to stretch it an additional 0.072 m? 1. 0.420 J 2. 0.486 J 3. 0.552 J 4. 0.398 J 5. 0.376 J 6. 0.464 J 7. 0.574 J

Answer 1

Work done is 0.442J

Explanation:

Given:

Spring constant, k = 33 N/m

Distance, x₁ = 0.15m

Additional distance, x₂ = 0.072 m

Total distance = 0.15 + 0.072 m

                        = 0.222 m

Work done, W = ?

We can calculate work done by the formula

$W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2$

On substituting the value we get:

$W = \frac{1}{2}k [(x_2)^2 - (x_1)^2]\\ \\W = \frac{1}{2} X 33[(0.222)^2 - (0.15)^2]\\ \\W = \frac{1}{2}X 33 [ 0.0493 - 0.0225]\\ \\W = 0.442 J$

Therefore, work done is 0.442J