Question

A string of length ‘L’ and force constant ‘K’ is stretched to obtain an extension ‘l’. It is further stretched to obtain an extension ‘l1’. The work done in second string stretching is
A) $\dfrac{1}{2}K{l_1}(2l + {l_1})$
B) $\dfrac{1}{2}Kl_1^2$
C) $\dfrac{1}{2}K({l^2} + l_1^2)$
D) $\dfrac{1}{2}K(l_1^2 - {l^2})$

Answer 1

Hint:The law of conservation of energy states that, ‘energy can neither be created nor be destroyed.’ Now the first law of thermodynamics is an extension of the law of conservation of energy and it states that work and energy are mutually interconvertible. To solve this problem we are going to use the same concept.

Complete step by step answer:
We have a string of length ‘L’ and its force constant is ‘K’. When the string is stretched to obtain an extension ‘l’, the potential or strain energy stored within the string is given by,
${E_1} = \dfrac{1}{2}K{l^2}$
Now, the string is again stretched to obtain an extension of ‘l₁’ so the total extension of the string will now be ‘l + l₁’. So, the potential or strain energy stored within the string due to second stretching is given by,
${E_2} = \dfrac{1}{2}K{(l + {l_1})^2}$
Now, by energy conservation law we know that the change in energy (here potential energy) is equal to the work done. Hence,
Work Done = Change in Potential Energy
$ W = \Delta E $
 $ W = {E_2} - {E_1}$
$\Rightarrow W = \dfrac{1}{2}K{(l + {l_1})^2} - \dfrac{1}{2}K{l^2}$
$\Rightarrow W = \dfrac{1}{2}K(2l{l_1} + l_1^2)$
$\Rightarrow W = \dfrac{1}{2}K{l_1}(2l + {l_1})$

Hence, option A is the correct answer option.

Note:The same method can be applied if the problem is given for a spring. Just like a string if a spring is given an extension of ‘x’ and it has a force or spring constant ‘k’, then the potential energy stored in the spring is given by $E = \dfrac{1}{2}K{x^2}$
The energy stored in an object when it is deformed or strained under the action of some external force is known as strain energy. Its unit is Joules or N-m.