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Calculate the work done to stretch an elastic spring by 40cm. If a force of 10N produces an extension of 4cm in it. U

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Answer:


20\; {\rm J}.

Step-by-step explanation:

The question has given the force it takes to stretch this spring by a particular length. With that information, the amount of work completed in order to stretch this spring by a different length can be found using the following steps:

  • Find the spring constant
    k of the spring from the restoring force and the corresponding displacement of the spring, and
  • Apply the equation for the elastic potential energy
    \text{EPE} of ideal springs to find the energy required to stretch the spring by the given distance.

Note the unit conversion during the calculations- all values should be measured in standard units, with distances measured in meters.

To find the spring constant
k of an ideal spring, divide the restoring force by the change in the length of the spring.


\displaystyle \text{spring constant} = \frac{\text{restoring force}}{\text{change in length}}.

In this question, it is given that an external force of
10\; {\rm N} is needed to increase the length of the spring by
x = 4\; {\rm cm} = 0.04\; {\rm m} (note the unit conversion.) In other words, the restoring force on the spring would be
10\; {\rm N}\! in response to a change in length of
0.04\; {\rm m}. The spring constant would be:


\begin{aligned}k &= \frac{(\text{restoring force})}{(\text{change in length})} \\ &= \frac{10\; {\rm N}}{0.04\; {\rm m}} \\ &= 250\; {\rm N\cdot m^(-1)}\end{aligned}.

For an ideal spring of spring constant
k, if the length of the spring is changed by
x, the elastic potential energy
\text{EPE} stored in that spring would be:


\displaystyle \text{EPE} = (1)/(2)\, k\, x^(2).

The spring constant in this question is
k = 250\; {\rm N\cdot m^(-1)}, whereas the displacement of the spring needs to be
x = 40\; {\rm cm} = 0.40\; {\rm m} (again, note the unit conversion.) The elastic potential energy stored in this spring would be:


\begin{aligned} \text{EPE} &= (1)/(2)\, k\, x^(2) \\ &= (1)/(2)\, (250\; {\rm N\cdot m^(-1)})\, (0.40\; {\rm m})^(2) \\ &= 20\; {\rm N\cdot m} \\ &= 20\; {\rm J}\end{aligned}.

(Note that
1\; {\rm N\cdot m} = 1\; {\rm J}.)

In other words, when an external force stretches this spring by
0.40\; {\rm m} from the equilibrium position, the amount of work done on the spring would be
20\; {\rm J}.

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