Question

when a long spring is stretched by 2 cm, its potential energy is U. if the spring is stretched by 10 cm, the potential energy in it will be?
A. 2U
B. 25 U
C. U/5
D. 5U

Answer 1

We know that when a spring is stretched it gets elongated and the energy stored is called potential energy. The spring when extended a force is developed in it which has the tendency to bring back the spring to its original length. This force is called restoring force. Given us two different conditions and we need to find the potential energy is the second case. We can make two equations and divide them to solve.

Complete step by step answer:
For first case:
potential energy of the spring, $U=\dfrac{k{{x}^{2}}}{2}$, where k is the spring constant in of N/m and x is the displacement from the equilibrium position in metres.
x= 2cm= 0.02 m
potential energy,
$U=\dfrac{k\times {{0.02}^{2}}}{2} \\\ \therefore k=5000U \\\$
For second case:
x= 10 cm= 0.1 m
$U'=\dfrac{kx{{'}^{2}}}{2} \\\ \Rightarrow U'=\dfrac{k\times {{0.1}^{2}}}{2} \\\ \Rightarrow U'=\dfrac{k\times {{0.1}^{2}}}{2} \\\ \Rightarrow U'=\dfrac{5000U\times {{0.1}^{2}}}{2} \\\ \therefore k=25U$

So, the correct option is B.

Additional Information:
When force is applied to a material, we know that it either stretches or compresses in response to the applied force. In mechanics, the force applied per unit area is known as stress and is denoted by the symbol σ. The extent to which the material compresses or stretches is known as strain. Different materials respond differently to applied stress. This information is necessary for engineers while selecting materials for their structures.Hooke’s law states that the strain of the material is proportional to the applied stress within the elastic limit of that material.

Note: When we are calculating the potential energy of the spring, we need to keep in mind that both the spring constant and the displacement from the equilibrium position are to be taken in standard SI units and thus the energy comes out to be Joules, J.Here, we first had calculated the spring constant from the first condition and then used it in the second condition.