HOOKE’S LAW
the Maths of stretching
springs.
&nb=
sp;
When a spring (or wire) is stretched elastically, i.e. it returns fully =
to its
unstretched length, it follows Hooke’s law=
, that
is
&=
nbsp; Force
applied =3D spring constant x extension&nbs=
p; &=
nbsp; F
&=
nbsp; ( N ) =
; (
N/m or N/cm) ( m or cm =
) =
k x e=
This
means that every individual spring will stretch in a predictable way up its=
elastic limit.<=
/p>
1. =
A spring has unstretched length of 15cm. A force of 2.5 N stretches=
it
to 21cm. Calculate
a) The extension with =
2.5 N
=
&nb=
sp; =
b) The spring constant=
&=
nbsp; &nbs=
p; &=
nbsp; c)
The expected extension with 3.75 N
&=
nbsp; &nbs=
p; &=
nbsp; d)
The length with 3.75 N
2. =
A spring is used in weighing scales with a spring constant of =
25 N/cm<=
span
style=3D'font-size:15.0pt;mso-bidi-font-size:10.0pt'>
a) What is the extension =
with
50 N loaded on it?
b) What is its original l=
ength
if 90 N make it stretch to 8.4 cm?
c) What force would make =
it 7.2
cm long?
3. =
Car suspension shocks have springs with k =3D 12000 N/m that have a=
maximum
contraction from 50cm to 32cm.
a) What force would cause=
this?
b) What is the contraction
length with 1200 N?
c) Why would the shocks b=
reak
with a force of 5kN?
Physics For You p79 q=
4 q5 and p148 q1 q2 q3=
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