The question is incomplete; the complete question is;
Suppose a force of 40 N is required to stretch and hold a spring 0.2m from its equilibrium position.
a. Assuming the spring obeys Hooke's law, find the spring constant k.
b.How much work is required to compress the spring 0.1m from its equilibrium position?
c.How much work is required to stretch the spring 0.4 m from its equilibrium position?
d.How much additional work is required to stretch the spring 0.2m if it has already been stretched 0.2m from its equilibrium?
Answer:
a) 200 Nm-1
b) 1 J
c) 16 J
d) 12 J
Step-by-step explanation:
From;
F= Ke
Where;
K= force constant
e= extension
Hence K= F/e = 40 N/0.2 m = 200 Nm-1
b) W= 1/2 Ke^2
Since e= 0.1 m
W= 0.5 × 200 × (0.1)^2
W= 1 J
c)
W= 0.5 × 200 × (0.4)^2
W= 16 J
d)
Work in stretching by 0.2 m = W= 0.5 × 200 × (0.2)^2 = 4 J
If stretched an additional 0.2 m, we have a total of 0.4 m for which the energy is 16 J.
The additional energy now is; 16 J - 4 J = 12 J