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###### springs question

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1 year ago

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)

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FPhysics%2FA-level%2FTopic-Qs%2FAQA%2F04-Mechanics-and-Materials%2FSet-M%2FForce%2C%2520Energy%2520%26%2520Momentum%2520(Multiple%2520Choice)%25202%2520QP.pdf

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)

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FPhysics%2FA-level%2FTopic-Qs%2FAQA%2F04-Mechanics-and-Materials%2FSet-M%2FForce%2C%2520Energy%2520%26%2520Momentum%2520(Multiple%2520Choice)%25202%2520QP.pdf

Reply 1

1 year ago

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.

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)

Reply 2

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.

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!

Reply 3

1 year ago

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

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

Reply 4

1 year ago

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

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!

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