If a spring extends by x on loading, then the energy stored by the spr...
As the block is being pulled the tension developed in the spring is nothing but spring force. So, T = kx .Also, the energy developed while pulling the spring is, E = (kx^2)/2On multiplying and dividing the above relation with k we get,E = (k^2x^2)/2k = (kx)^2/2kTherefore we get the relation as, E = T^2/2k
If a spring extends by x on loading, then the energy stored by the spr...
Explanation:
When a spring is loaded or stretched by a certain distance, it stores potential energy within it. This potential energy is given by the formula:
Energy stored = (1/2)kx^2
Where:
- k is the spring constant (a measure of the stiffness of the spring)
- x is the distance the spring is stretched or extended
Now, let's understand why the correct answer is option 'B' - T^2/2k.
1. Relationship between tension (T) and spring constant (k):
The tension (T) in a spring is directly proportional to its spring constant (k). This relationship is given by Hooke's Law:
T = kx
Where:
- T is the tension in the spring
- k is the spring constant
- x is the distance the spring is stretched or extended
2. Substituting the value of T in the energy stored formula:
Using the relationship T = kx, we can substitute the value of T in the energy stored formula:
Energy stored = (1/2)(T^2/k)
3. Simplifying the expression:
We can simplify the expression by multiplying T^2/k with 1/2:
(1/2)(T^2/k) = T^2/2k
4. Conclusion:
Therefore, the energy stored by the spring when it extends by x on loading is T^2/2k. This means that the correct answer is option 'B'.