# When two springs in parallel with a spring constant of 80 N/m and 120 N/m are pulled by a force of 18 N over 8 cm, what is the energy required.

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### 1 Answer

When two springs with spring constants k1 and k2 are connected in parallel and pulled by a force F, the force is divided between the two and is equal to F*k1/(k1 + k2) and F*k2/(k1 + k2) respectively. The application of the force changes the length of each spring by the same extent.

In the problem the spring constant of the springs in parallel is 80 N/m and 120 N/m and a force of 18 N is applied over 8 cm. The increase in energy of the spring with a spring constant 80 N/m is 18*(80/200)*(8/100) and the increase in energy of the spring with a spring constant 120 N/m is 18*(120/200)*(8/100)

The total energy required is the sum of the increase in energy in each spring and equal to 18*(80/200)*(8/100) + 18*(120/200)*(8/100) = 18*8/100 = 1.44 J

It should be noticed here that the calculation of the total energy required is done by multiplying the force by the distance over which it is applied; the spring constants do not have any relevance in the problem. They would have been required if the energy stored in each spring had to be determined.

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