Question

If a force of 200 [N] is exerted on a spring whose length is 25 [cm], and it is observed that the spring length reaches a value of 30 [cm], what is the spring constant of the spring?

A. 4000 N/m.
B. 1500 N/m.
C. 3000 N/m.
D. 2000 N/m.

Answer 1

The spring constant of the spring is 4000 N/m.

Step-by-step explanation:

The spring constant of a spring can be calculated using Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The formula for the spring constant is:

k = F / x

where k is the spring constant, F is the force exerted on the spring, and x is the displacement of the spring.

In this case, a force of 200 [N] is exerted on a spring whose length increases from 25 [cm] to 30 [cm]. The displacement, x, can be calculated as:

x = 30 [cm] - 25 [cm] = 5 [cm]

Substituting the values into the formula:

k = 200 [N] / 5 [cm]

Converting cm to meters:

k = 200 [N] / 0.05 [m] = 4000 [N/m]

Therefore, the spring constant of the spring is 4000 N/m. So, the correct answer is A. 4000 N/m.