Question

the potential energy of a long spring when stretched by 2 centimetre is U .if the spring is stretched by a 8 cm the potential energy stored in this is

Answer 1

U= Initial potential energy. U'= Final Potential Energy. K=Spring Constant

{0.5 k (0.08)^2} / {0.5 k (0.02)^2} = U' / U
=> New Potential Energy (U')= 16 U

Answer 2

Potential energy of a spring = 1/2×Kx^2, where K is the spring constant and x is the length stretched or compressed.
Here, K has to be found and x = 2cm.
Potential energy of spring = U
Therefore, U = 1/2×K×(2)^2.
Implies, U = 2K.
Hence, K = U/2.
Hence, we have found the value of the spring constant.
Now, we know the value of the spring constant and the extent to which it is stretched, 8cm.
We are supposed to find out the new potential energy.
Therefore, new potential energy = 1/2×U/2×(8)^2 = 16U.


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