Question

A spring with a spring constant of $k = 200 \, \text{N/m}$ is stretched by an external force of $F_{\text{ext}} = 10 \, \text{N}$. The spring elongates by a distance of $x = 0.5 \, \text{m}$. Calculate the following:

a) The potential energy stored in the spring.

b) The elastic potential energy when the spring is stretched further by an additional distance of $0.3 \, \text{m}$.

c) The work done by the external force to stretch the spring by the additional $0.3 \, \text{m}$.

Answer

a) The potential energy stored in a spring can be calculated using the formula: [ PE_{\text{spring}} = \frac{1}{2} k x^2 ] where $k$ is the spring constant and $x$ is the displacement from the equilibrium position.

Substituting the given values: [ PE_{\text{spring}} = \frac{1}{2} (200 , \text{N/m}) (0.5 , \text{m})^2 ] [ PE_{\text{spring}} = \frac{1}{2} (200 , \text{N/m}) (0.25 , \text{m}^2) ] [ PE_{\text{spring}} = 25 , \text{J} ]

b) To calculate the elastic potential energy when the spring is stretched further, we need to determine the new displacement $x'$.

Given the additional distance stretched as $0.3 \, \text{m}$, the new displacement is: [ x' = x + \Delta x = 0.5 , \text{m} + 0.3 , \text{m} = 0.8 , \text{m} ]

Using the same formula as before: [ PE_{\text{spring}}' = \frac{1}{2} (200 , \text{N/m}) (0.8 , \text{m})^2 ] [ PE_{\text{spring}}' = \frac{1}{2} (200 , \text{N/m}) (0.64 , \text{m}^2) ] [ PE_{\text{spring}}' = 51.2 , \text{J} ]

c) The work done by an external force to stretch the spring can be calculated using the formula: [ \text{Work} = \frac{1}{2} k (x'^2 - x^2) ] where $x'$ is the final displacement and $x$ is the initial displacement.

Substituting the given values: [ \text{Work} = \frac{1}{2} (200 , \text{N/m}) [(0.8 , \text{m})^2 - (0.5 , \text{m})^2] ] [ \text{Work} = \frac{1}{2} (200 , \text{N/m}) (0.64 , \text{m}^2 - 0.25 , \text{m}^2) ] [ \text{Work} = \frac{1}{2} (200 , \text{N/m}) (0.39 , \text{m}^2) ] [ \text{Work} = 39 , \text{J} ]

Therefore, the answers are:

a) The potential energy stored in the spring is $25 \, \text{J}$.

b) The elastic potential energy when the spring is stretched further by an additional distance of $0.3 \, \text{m}$ is $51.2 \, \text{J}$.

c) The work done by the external force to stretch the spring by the additional $0.3 \, \text{m}$ is $39 \, \text{J}$.