Question

The amount of work done in stretching a spring of force constant \[500N/m\] from a stretched length of \[40cm\] to \[50cm\]is
(A) \[45J\]
(B) \[22.5J\]
(C) \[45 \times {10^4}J\]
(D) \[22.5 \times {10^4}J\]

Answer 1

Hint: When the spring is stretched by an external force, its length is changed, this length is called stretched length.
Use the relation between Work-done by a spring with the extended length of the spring and calculate the amount of work done using the change of lengths and force constant.
Use the two values of stretched lengths given in the problem in the relation for the same values of the force and the force constant.

Formula used:
If the spring is stretched by the force applied externally,
We can write that the work-done by the spring is, \[W = \dfrac{1}{2}k{x^2}\]
Where \[k\]=The force constant of the spring,
\[x\]= the extended or stretched of the spring.

Complete step by step answer:
An external force is applied to a spring. The length of the spring is extended to a certain amount due to this external force.
The force that is applied to a spring is directly proportional to the extended length.
If the applied force is \[F\] and the extended or stretched length of the spring is \[x\], then \[F\propto x\]
\[\therefore F = kx\], where \[k\]=The force constant of the spring,
Now the amount of the work done is the product of the average force and the displacement.
Here the displacement is taken as the amount of the extended length of the spring.
So we may write, \[W = \dfrac{F}{2} \times x\]
Since, \[F = kx\]
\[ \Rightarrow W = \dfrac{1}{2}k{x^2}\]
In the problem, during two times two stretched lengths of the spring are given where the amount of applied force is the same.
So we may rewrite the equation \[W = \dfrac{1}{2}k{x^2}\]as,
\[W = \dfrac{1}{2}k({x_2}^2 - {x_1}^2)\], where \[{x_1}\] is the initial stretched length and \[{x_2}\] is the final stretched length.
Given, \[k\]= \[500N/m\]
\[{x_1}\]=\[40cm\]
\[ \Rightarrow {x_1} = \dfrac{{40}}{{100}}m = \dfrac{2}{5}m\]
and \[{x_2}\]=\[50cm\]
\[ \Rightarrow {x_2} = \dfrac{{50}}{{100}}m = \dfrac{1}{2}m\]
\[\therefore W = \dfrac{1}{2}500 \times \left[ {{{(\dfrac{1}{2})}^2} - {{(\dfrac{2}{5})}^2}} \right]\]
\[ \Rightarrow W = \dfrac{1}{2}500 \times \left[ {(\dfrac{1}{4}) - (\dfrac{4}{{25}})} \right]\]
\[ \Rightarrow W = \dfrac{1}{2}500 \times \left[ {\dfrac{{25 - 16}}{{100}}} \right]\]
\[ \Rightarrow W = \dfrac{1}{2} \times 5 \times 9\]
\[ \Rightarrow W = 22.5J\]
Work-done by the spring is, \[W = 22.5J\]

Hence, the right answer is in option (B).

Note:
It is very important to get that the value of the \[x\] is not the original length of the spring.it is the extended portion of the spring which is occurring by the applied force.
The displacement is taken for calculating the force as well as the amount of work done is the length of the extended portion since we take such as when the force is not applied i.e. \[F = 0\] there is no extension i.e \[x = 0\].
The length of the spring can be also extended by its weight when it is hanging.