Question

Use hookes law
A force of 265 newtons stretches a spring 0.15 meters a) what force stretches the spring 0.2
B)how far does a force of 140 Newton’s stretch the spring

Answer 1

we know that

Hooke's law

$F=kx$

where

F is force

k is spring constant

x is stretched length

we are given F=265N and x=0.15m

now, we can use it and find k

$265=k*0.15$

we get

$k=\frac{5300}{3}$

now, we can plug back

$F=\frac{5300}{3}x$

(A)

we are given

stretches the spring 0.2

so, x=0.2

and then we can find F

$F=\frac{5300}{3}*0.2$

$F=353.333N$

(B)

we are given

F=140N

so, we can set it equal and then we can solve for x

$140=\frac{5300}{3}*x$

$x=0.07925m$............Answer