# springs question

I keep getting D for this question, but the answer is B. Can someone show me how to work out the answer 0.2?

A load of 4.0 N is suspended from a parallel two-spring system as shown in the diagram.
The spring constant of each spring is 20 N mâ€“1.
The elastic energy, in J, stored in the system is
A 0.1
B 0.2
C 0.4
D 0.8

Diagram reference (question 10)
When springs are connected in parallel, their effective spring constant is K1 + K2. So in this case, its 20 + 20 = 40 N m-1
F = KL
4 = 40L
L = 4/40 = 0.1

Energy = 1/2 x F x L = 1/2 x 4 x 0.1 = 0.2 Joules

When springs are connected in series, their effective spring constant: 1/K(total) = 1/K1 + 1/K2

Its similar to the resistance equations but in reverse.
(edited 1 year ago)
Original post by speed_bird
When springs are connected in parallel, their effective spring constant is K1 + K2. So in this case, its 20 + 20 = 40 N m-1
F = KL
4 = 40L
L = 4/40 = 0.1

Energy = 1/2 x F x L = 1/2 x 4 x 0.1 = 0.2 Joules

When springs are connected in series, their effective spring constant: 1/K(total) = 1/K1 + 1/K2

Its similar to the resistance equations but in reverse.

Oh I realised I mixed it up with resistance. I get it now. Thank you!
An alternative approach without using effective k.

You can consider for 1 spring. Since there are 2 springs, each one has a spring force of 2 N. Using the k value, can deduce the extension = 0.10 m.

Energy stored in one spring is 1/2 k x^2 = 0.10 J.

With 2 springs, the total energy stored will be 0.20 J
Original post by Ethanity007
An alternative approach without using effective k.

You can consider for 1 spring. Since there are 2 springs, each one has a spring force of 2 N. Using the k value, can deduce the extension = 0.10 m.

Energy stored in one spring is 1/2 k x^2 = 0.10 J.

With 2 springs, the total energy stored will be 0.20 J

That makes sense as well. Thank you!