Question

A weight of 8.7 N is attached to a spring that has a spring constant of 190 N/m. How much will the spring stretch?

Answer 1

Given:
Fs=8.7N
K=190N/m
Unkown:
X=?
Equation:
Fs=Kx
8.7N=(190N/m)x
X=4.6*10power of-2m

Answer 2

The spring will stretch "4.6×10⁻2 m".

Given values are:

• Force, Fs = 8.7 N
• Spring constant, k = 190 N/m

As we know the relation,

→ $Fs = K\times x$

or,

→   $x = \frac{Fs}{k}$

By putting the values, we get

        $= \frac{8.7}{190}$

        $= 4.6\times 10^{-2} \ m$

Thus the response above is right.

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