Question

Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.46 kg mass to the spring’s other end. The acceleration of gravity is 9.81 m/s 2 . If the spring stretches 3.7 cm from its equilibrium position, what is the spring constant?

Answer 1

Answer: The value of spring constant is 121.9 N/m

Explanation:

Force is defined as the mass multiplied by the acceleration of the object.

$F=m\times g$

where,

F = force exerted on the object  = ?

m = mass of the object  = 0.46 kg

g = acceleration due to gravity = $9.81m/s^2$

Putting values in above equation, we get:

$F=0.46kg\times 9.81m/s^2=4.51N$

To calculate the spring constant, we use the equation:

$F=k\times x$

where,

F = force exerted on the spring = 4.51 N

k = spring constant = ?

x = length of the spring = 3.7 cm = 0.037 m     (Conversion factor:  1 m = 100 cm)

Putting values in above equation, we get:

$4.51N=k\times 0.037m\\\\k=\frac{4.51N}{0.037m}=121.9N/m$

Hence, the value of spring constant is 121.9 N/m