9702_s22_qp_12 Question 19

A metal wire obeys Hooke’s law and has a Young modulus of 2.0 x 10¹¹ Pa. The wire has an original length of 1.6 m and a diameter of 0.48 x 10^-3 m.

What is the spring constant of the wire?

A 7.2 x 10³ N m–1

B 2.3 x 10⁴ N m–1

C 2.9 x 10⁴ N m–1

D 9.0 x 10⁴ N m–1

ANSWER: B

The Physics Behind

• The spring constant can be derived from the spring version of the Hooke's Law, that is, force F is directly proportional to extension x as

k = F/x

• The other version of the Hooke's Law states that the stress [that is F/A] is directly proportional to strain [that is x/L]. The proportionality constant here is the Young modulus Y.

Y = F/A / x/L

• Rearranging the second equation above, you should see that

Y = FL / Ax

Y = (F/x) (L/A)

• Note that F/x is the spring constant k that you need. So,

Y = k (L/A)

k = AY/L

k = 1/4 πd²Y/L

• Substituting the given data, k is

k = ((0.25) π (0.48 x 10^-3 m)² (2.0 x 10¹¹ P))/ (1.6 m)

k = 23000 N m–1