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When two springs in parallel with a spring constant of 80 N/m and 120 N/m are pulled by...
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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.
Posted by justaguide on March 24, 2012 at 2:13 PM (Answer #1)
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