Question

The diagram shows a spring without a load and then with a load of mass 500 g suspended from the same spring. The spring obeys Hooke’s law.

The length of the unloaded spring is 30 cm.
When the 500 g load is suspended from the spring, the spring extends to a new length of 35 cm.
The gravitational field strength g is 10 N/ kg.
Which calculation gives the spring constant of the spring?

(A) \(\frac{0.5 \times 10}{35 – 30} N/cm\)
(B) \(\frac{0.5 }{10 \times (35-30)} N/cm\)
(C) \(\frac{10 \times (35-30)}{0.5} N/cm\)
(D)  \(0.5 \times 10 \times (35-30) N/cm\)