Question

8. The force constant of a spring is 150. N/m. (a) How much force is required to stretch the spring 0.25 m? (b) How much work is done on the spring in that case?

Answer 1

Answer:

a) 37.5N

b) 9.375Joules

Explanation:

a) According to Hooke's law

F = ke

k is the spring constant

e is the extension;

F = 150 * 0.25

F = 37.5N

b) Work done on the spring = 1/2ke^2

Work done on the spring = 1/2 * 150 * 0.25^2

Work done on the spring  = 75 * 0.0625

Work done on the spring  = 9.375Joules