Question

1. When the spring is stretched by 5
cm, applying a force of 2.5 Nm , what
is the spring constant ? How will the
spring be stretched by force of 2 N
​

Answer 1

Explanation:

Motion of a spring is a case of simple harmonic motion.

In S. H. M,

F = -kx

where F is the force applied;k is the spring constant;and x is the change in natural length of spring.

Using the equation,

2.5 = -0.5k

-k=2.5/0.5

k = -5N/m^2

When 2N force is applied,

2 = -5x

x = -0.4m Or 40 cm