# Fining the force constant of a spring

Crbeach
When a 9.09kg mass is placed on top of a vertical spring, the spring compresses 4.18 cm. Find the force constant of the spring.

Okay I know the formula for a spring is F=-kx
So k=mg/x
converting the cm to m I get 4.18cm = 0.0418m

So I plug in my numbers
k=(9.09kg)(9.81m/s^2)/ 0.0418
but that gives me 2133 which is not the answer.

What am I doing wrong?

## Answers and Replies

5carola5
when I type it in my calculator I get:
(9,09 * 9,81) / (4,18^(-2)) = 1 558,06458

If that is the right answer than you just made a typo, if it's not then I also don't know:P

Homework Helper
Include the unit of the spring constant.

ehild

Crbeach
The answer is 2.13 kg*m/s^2 / m

The only way I can see to get that answer is to make the cm into decimeters
4.18 cm = 41.8 dec
That gives me 2.13 but that wouldn't work for the unit kg*m/s^2 / m

Last edited:
Homework Helper
4.18 cm is 0.418 dm...

Your answer k=2133 N/m is correct if the mass is 9.09 kg. If the mass is 9.09 g (gram) k would be 2.13 N/m.

ehild

Crbeach
Ok so there is a typo in the book, either in the question or in the answer.
Thank you very much. Its nice to know I was doing the problem correctly

Dansar
I'm guessing that you were off by a factor of 2. In the Hooke's Law equation (F = -kx) where force is generated by gravity, F represents the AVERAGE force, not the peak force. Force is not constant as the the spring is compressed from its relaxed length to its new equilibrium point with the weight sitting on it.

The problem assumes that the spring starts from the relaxed length with zero force. The force at the maximum compression is the peak force (mg). Since k is a constant, the relationship between force and distance is linear. So, the average force is 1/2 of the peak force (1/2 * m * g).

Homework Helper
I'm guessing that you were off by a factor of 2. In the Hooke's Law equation (F = -kx) where force is generated by gravity, F represents the AVERAGE force, not the peak force.

That is wrong.
F=-kx is the force the spring exerts on the block when its length differs by x from the relaxed length, not an "average" one. The block is stationary, so the force of gravity is cancelled by the spring force, mg-kx=0.

ehild