Question

When a spring is stretched by 2 cm, it stores 100 J of energy. If it is stretched further by 2 cm, the stored energy will be increased by
(A). 100 J
(B). 200 J
(C). 300 J
(D). 400 J

Answer 1

Hint: To solve this question find the mathematical expression for the energy stored in a spring. Find the value of spring constant by using the given values in the first condition. Then find the value of energy stored for the second condition and subtract the first value to get the final answer.

Formula used:
$E=\dfrac{1}{2}k{{x}^{2}}$
Where, E is the energy stored in the spring, k is the spring constant and x is the displacement of the spring from its equilibrium position.

Complete step by step answer:
The potential energy of a spring or the energy stored in a spring is given by the equation,
$E=\dfrac{1}{2}k{{x}^{2}}$
Where, E is the energy stored in the spring, k is the spring constant and x is the displacement of the spring from its equilibrium position.
When the spring is stretched by 2 cm (=0.02 m), the stored energy is 100 J.
We can find the value of spring constant k by putting the given values on the above equation.
$\begin{aligned} & {{E}_{1}}=\dfrac{1}{2}k{{x}^{2}}=100\text{ J} \\\ & \dfrac{1}{2}k\times {{0.02}^{2}}=100\text{ J} \\\ & \dfrac{1}{2}k\times 0.0004=100\text{ J} \\\ & k=\dfrac{100\times 2}{.0004}N{{m}^{-1}} \\\ & k=500000N{{m}^{-1}} \\\ \end{aligned}$
So, value of the spring constant k is, $k=500000N{{m}^{-1}}$
Now, if we stretch the spring further by 2 cm, x will be 4 cm (0.04 m).
Hence, putting these values on the equation again,
$\begin{aligned} & {{E}_{2}}=\dfrac{1}{2}k{{x}^{2}} \\\ & {{E}_{2}}=\dfrac{1}{2}\times 500000\times {{0.04}^{2}} \\\ & {{E}_{2}}=250000\times 0.0016 \\\ & {{E}_{2}}=400J \\\ \end{aligned}$
So, the energy stored will be increase by
$\begin{aligned} & {{E}_{2}}-{{E}_{1}} \\\ & =400J-100J \\\ & =300J \\\ \end{aligned}$
So, the correct option is (c).

Additional information:
Hooke's law gives the relation between the restoring force and the displacement of the spring from its equilibrium position.
${{F}_{s}}=-kx$

Note: Always check if the units of all the physical quantities involved in an equation is the same or not. The spring constant is also known as the stiffness constant. The spring constant can be defined as the force required to stretch or compress the spring by a unit length.