Question

A stretched spring has an elastic potential energy of 35 J when it is stretched 0.54 m. What is the spring constant of the spring? Round your answer to two significant figures.
StartFraction N over m EndFraction

Answer 1

Answer:

240 N/m

Explanation:

Using the formula as follows:

U = ½kx²

Where;

U = elastic potential energy (J)

K = spring constant (N/m)

x = stretched displacement (m)

According to the information provided in this question,

U = 35J

x = 0.54m

k = ?

U = ½kx²

35 = ½ × k × 0.54²

35 = ½ × k × 0.2916

35 = 0.2916k/2

70 = 0.2916k

k = 70 ÷ 0.2916

k = 240.05

To 2s.f, the spring constant (k) = 240 N/m.