
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
570.3k+ views
Hint: 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).
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).
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

