Question

When you stretch a spring 13 cm past its natural length, it exerts a force of 21 N. What is the spring constant of this spring?

Answer 1

Answer:

Explanation:

From Hooke's law, $F=-kx$

Substistuting the values, $21N= -k 13cm$ you can ignore the minus sign, since it is there to state that the force is always in the opposite direction of the deformation, and has no influence otherwise, and your constant is $1,6154 *10-2 N/m$

Answer 2

Answer:

Spring constant, k = 161.5 N-m

Explanation:

It is given that,

Force on spring, F = 21 N

The spring is stretched 13 cm past its natural length i.e. x = 0.13 m

The Hooke's law gives the relationship between the force and the compression in the spring. Mathematically, it can be written as :

$F=-kx$

k is spring constant

-ve sign shows the direction of force is opposite

So, $k=\dfrac{F}{x}$

$k=\dfrac{21\ N}{0.13\ m}$

k = 161.5 N-m

Hence, this is the required solution.