patysando92
Answered

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 :

skyluke89
Hook's law states that:
[tex]F=kx[/tex]
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
[tex]x=6 cm = 0.06 m[/tex]
while the force applied is the weight of the mass m=200 g=0.2 kg hanging on the spring:
[tex]F=mg=(0.2 kg)(9.81 m/s^2)=1.96 N[/tex]
So, we can find the constant of the spring by using Hook's law:
[tex]k= \frac{F}{x}= \frac{1.96 N}{0.06 m}=32.7 N/m [/tex]
hamzaahmeds

This question involves the concept of Hooke's Law.

The value of the spring constant is "32.7 N/m".

HOOKE'S LAW

According to Hooke's Law, the change in length of a spring is directly proportional to the force applied to it. Mathematically,

[tex]F=kx\\\\k =\frac{F}{x}[/tex]

where,

  • k = spring constant = ?
  • F = force applied = weight = mg = (0.2 kg)(9.81 m/s²) = 1.962 N
  • x = change in length = 6 cm = 0.06 m

Therefore,

[tex]k=\frac{1.962\ N}{0.06\ m}[/tex]

k = 32.7 N/m

Learn more about Hooke's Law here:

https://brainly.com/question/3355345

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