Question

A spring of natural length 1.5m is extended 0.005m by a force of 0.8N.What will its length be when the applied force is 3.2N​

Answer 1

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