find the energy

Question

Use g=10ms−2. A 2 kg mass is hung from an ideal spring with spring constant 40 N/m. The spring is then pulled down 20 cm from its equilibrium position. The work done in J due to the extension from equilibrium is closest to:

Answer 1

Work done = \(\int \vec F \cdot d\vec r\)

Let y positive direction be pointing towards ground. 

Then:

\(\vec F_1=-ky\hat j\)

\(\vec F_2 = mg \hat j\)

\(d\vec r = dy \hat j\)

\(20cm = 0.2m\)

Work is then \(\int_0^{0.2}mgdy - \int_0^{0.2}kydy\) =  \(0.2m \cdot 10\frac{m}{s^2}\cdot 2kg - 40\frac{N}{m}\cdot(0.2m)^2\cdot 0.5 = 4J - 0.8J = 3.2J\)