Question

a spring is stretched 6 cm when mass of 200g is hung on it calculate the spring constant of this spring plz help it's for my homework!!!

Answer 1

Hook's law states that:
$F=kx$
where F is the force applied on the spring, k is the spring constant and x the stretch/compression of the spring.
In our problem, the stretch is
$x=6 cm = 0.06 m$
while the force applied is the weight of the mass m=200 g=0.2 kg hanging on the spring:
$F=mg=(0.2 kg)(9.81 m/s^2)=1.96 N$
So, we can find the constant of the spring by using Hook's law:
$k= \frac{F}{x}= \frac{1.96 N}{0.06 m}=32.7 N/m$