Question

A weight of 10 N hangs motionless from a spring. The spring is stretched 0.286 m from its original length. What is the force constant for the spring?

a-2.8 N/m

b-22 N/m

c-35 N/m

d-43 N/m

Answer 1

C. -35N/m

Step-by-step explanation:

Using Hooke's Law, we know that the elastic constant of a spring can be found with this formula:

`F=-kx`

where:

`F` is the force ( in Newtons)

`k` is the constant ( in Newtons per meter)

`x` is the extension ( in meters)

thus, by rearranging the formula we can get:

`k=-(F)/(x)`

`k = -(10)/(0.286)`

`k=-34.96503N/m`

Rounding off the answer, we get -35N/m