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anonymouss2004
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#1
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 m–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...202%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 m–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...202%2520QP.pdf
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username5318474
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#2
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#2
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.
Last edited by username5318474; 1 month ago
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anonymouss2004
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#3
(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.
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.
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Ethanity007
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#4
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#4
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
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anonymouss2004
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#5
(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
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
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