Natural length of spring
l
=
1.5
m
Extension of spring
x
1
=
0.005
m
Force acting on spring
F
1
=
0.8
N
Force acting on spring
F
2
=
3.2
N
x
2
be the new extension of spring
According to Hooke's law, we have;
F
=
K
x
F
1
F
2
=
x
1
x
2
x
2
=
F
2
x
1
F
1
x
2
=
3.2
×
0.005
0.8
x
2
=
0.02
m
Therefore the new length of the spring is,
l
n
e
w
=
l
+
x
2
l
n
e
w
=
1.5
+
0.02
l
n
e
w
=
1.52
m