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?

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

Answers

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 NSpring constant, k = 190 N/m

As we know the relation,

→ [tex]Fs = K\times x[/tex]

or,

→   [tex]x = \frac{Fs}{k}[/tex]

By putting the values, we get

        [tex]= \frac{8.7}{190}[/tex]

        [tex]= 4.6\times 10^{-2} \ m[/tex]

Thus the response above is right.

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