A spring has a constant of 20 n/m. how far will it extend when the following forces are applied. (assume the spring does not pass its elastic limit)

a) 10 N

b) 50 N

c) 0.2 KN

A.) 0.5 m

B.) 2.5 m

C.) 10 m

Explanation:

Given that the

Spring constant K = 20 N/m

From hooks law

F = Ke

Where

F = force applied

K = spring constant

e = extension

A.) When F = 10 N

Substitute F and K into the formula

10 = 20e

e = 10/20 = 0.5 m

B.) When F = 50 N

50 = 20e

e = 50/20 = 2.5 m

C.) When F = 0.2 KN

0.2 * 1000 = 20e

200 = 20e

e = 200/20 = 10m