Question

Multiple Choice A spring has a natural length of 0.10 \(\mathrm{m}\) . \(\mathrm{A} 200\) -n force stretches the spring to a length of 0.15 \(\mathrm{m}\) . Which of the following gives the work done in stretching the spring from 0.10 \(\mathrm{m}\) to 0.15 \(\mathrm{m} ?\) (A) 0.05 \(\mathrm{J} \quad\) (B) 5 \(\mathrm{J} \quad\) (C) 10 \(\mathrm{J}\) (D) 200 \(\mathrm{J} \quad\) (E) 4000 \(\mathrm{J}\)

Short answer

The correct choice is (B) 5 Joules.

Step-by-step solution

Calculate the spring constant

Using Hooke's Law \(F = kx\), we can Rearrange equation to find k, which is given as: \(k = F/x\). Substituting the given values: \(k = 200\, N/0.05\, m = 4000\, N/m\).

Find the work done

The work done to stretch or compress a spring is given by the equation \(W = 0.5kx^2\), where \(x\) is the amount the spring is stretched or compressed from its natural length. Substituting the calculated value of \(k\) and the given stretch of the spring: \(W = 0.5 * 4000\, N/m * (0.05\, m)^2 = 5\, J\).

Match the result with the options

The calculated work done is 5 Joules, which matches with option (B). So, the correct choice is (B) 5 Joules.