Question

4J of work is required to stretch a spring through 10 cm beyond its unstretched length. The extra work required to stretch it through additional 10 cm shall be
A. 4J
B. 8J
C. 12J
D. 16J

Answer 1

In this problem, extra work required to be done is asked. So, to find that you have to find the spring constant first. Use the formula for work done, substitute the values for work done required to stretch initial 10cm and calculate the spring constant. Now, add the additional stretched distance to the initial distance and find the total work done. Then, subtract the value of work done for initial 10 cm from the total work done. This will give the value for extra work done required to stretch the additional distance.

Formula used:
$W= \dfrac {1}{2}k{x}^{2}$

Complete answer:
Given: Amount of work done to stretch the spring (W) = 4J
Distance (x)= 10 cm= 0.1m
Work done is given by,
$W= \dfrac {1}{2}k{x}^{2}$
Where, k is the spring constant
Substituting the given values in above equation we get,
$4= \dfrac {1}{2} \times {0.1}^{2}$
$\Rightarrow k= \dfrac {4 \times 2}{0.01}$
$\Rightarrow k= 800{N}/{m}$
The spring is stretched by additional 10cm, therefore
${x}_{1}= 10+10$
$\therefore {x}_{1}= 20cm$
$\Rightarrow {x}_{1}= 0.2 m$
Now, the total work done to stretch the spring by 20cm is given by,
$W= \dfrac {1}{2}k{x}_{1}^{2}$
Substituting the values in above equation we get,
$W= \dfrac {1}{2}\times 800 \times {0.2}^{2}$
$\Rightarrow W= \dfrac {1}{2} \times 32$
$\Rightarrow W= 16J$
Thus, the extra work required to be done will be $16J-4J=12J$.

So, the correct answer is “Option C”.

Note:
When a spring is compressed or relaxed, a restoring force is stored in it which is proportional to the displacement of the spring. The spring constant defines the force which is required to stretch or compress the spring by a unit length. It is also known as stiffness constant. As mentioned above, additional work can be calculated by subtracting the initial work from total work or the final work. We know, work is equal to energy, therefore their units are same i.e. Joule (J).