Question

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

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 stretched length of the spring and calculate the amount of work done using the change of lengths and force constant.
Put the two values of stretched lengths given in the problem for the same values of force and force constant.

Formula used: When the spring is stretched by an external force, the work done by the spring, \[W = \dfrac{1}{2}k{x^2}\]
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 and that’s why the length of the spring is extended to a certain amount. The force 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 defined as the product of the average force and displacement.
So we may write, \[W = \dfrac{F}{2} \times x\]
Since, \[F = kx\]
\[W = \dfrac{1}{2}k{x^2}\]
In the problem, two stretched lengths of the spring are given for two moments but 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.
It is given, \[k = 500N/m\]
\[{x_1}\]=\[40cm\]
\[{x_1} = \dfrac{{40}}{{100}}m\]
On dividing the term and we get,
\[ \Rightarrow \dfrac{2}{5}m\]
and \[{x_2} = 50cm\]
\[{x_2} = \dfrac{{50}}{{100}}m\]
Let us divide the terms we get,
\[ = \dfrac{1}{2}m\]
Now we have to put the values in the formula and we get,
\[\therefore W = \dfrac{1}{2}500 \times \left[ {{{(\dfrac{1}{2})}^2} - {{(\dfrac{2}{5})}^2}} \right]\]
Taking squaring the bracket terms and we get.
\[ \Rightarrow W = \dfrac{1}{2}500 \times \left[ {(\dfrac{1}{4}) - (\dfrac{4}{{25}})} \right]\]
Let us take the LCM of the bracket term and we get
\[ \Rightarrow W = \dfrac{1}{2}500 \times \left[ {\dfrac{{25 - 16}}{{100}}} \right]\]
On subtracting the term and we get,
\[ \Rightarrow W = \dfrac{1}{2}500 \times \left[ {\dfrac{9}{{100}}} \right]\]
On cancel the term and we get,
\[ \Rightarrow W = \dfrac{1}{2} \times 5 \times 9\]
On simplifying the term and we get,
\[ \Rightarrow W = 22.5J\]
Work-done by the spring is, \[W = 22.5J\]

Hence, the right answer is in option \[\left( {\text{B}} \right)\].

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 spring may be extended by its weight while hanging it.