Question

The tension in a spring of spring constant k is T. The potential energy of the spring is:
(a) T²/k^2
(b) T²/k
(c)2T^2/k
(d) T^2/2k

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Answer 1

Answer:

Option d) T^2/2K.

Explanation:

Since, from the question we know that the tension in the spring is given as T and the spring is having the spring constant as K. So, we know that the small distance x moved by the spring is x=F/K.

Again, we know that the potential energy stored in the spring will be U =1/2Kx^2 in which substituting the value of x we will get that potential energy is 1/2K(F/K)^2 which on solving we will get T^2/2K.