91Ó°ÊÓ

The elastic potential energy of a spring stretched or compressed by a certain distance is equal to the work done by the external force in moving the spring by this distance starting from its equilibrium position.

If a spring of spring constant k is compressed or expanded by a distance x, then the elastic potential energy of the spring will be \(P{E_{{\rm{el}}}} = \frac{1}{2}k{x^2}\).

Spring is stretched from its equilibrium length by a distance:

\(3.0 \times {10^4}\;{\rm{N/m}}\)

The elastic potential energy of the spring, when stretched by a distance x, is \(P{E_{{\rm{el}}}} = 6.0\;{\rm{J}}\).

When spring is stretched 4.0 cm, the total distance of stretched spring from the equilibrium length is as follows:

\(\begin{array}{c}x' = \left( {2.0 + 4.0} \right)\;{\rm{cm}}\\ = 6.{\rm{0}}\;{\rm{cm}}\\ = 6.0\;{\rm{cm}} \times \left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)\\ = 6.0 \times {10^{ - 2}}\;{\rm{m}}\end{array}\).

Let spring constant of spring is k.

If the equilibrium position of the spring is taken as the reference position, the elastic potential energy of the spring when stretched by a distance x is given as follows:

\(\begin{array}{c}P{E_{{\rm{el}}}} = \frac{1}{2}k{x^2}\\6.0\;{\rm{J}} = \frac{1}{2}k{\left( {2.0 \times {{10}^{ - 2}}\;{\rm{m}}} \right)^2}\\k = \frac{{2 \times 6.0\;{\rm{J}}}}{{{{\left( {2.0 \times {{10}^{ - 2}}\;{\rm{m}}} \right)}^2}}}\\k = 3.0 \times {10^4}\;{\rm{N/m}}\end{array}\)

Thus, the spring constant of the spring is \(3.0 \times {10^4}\;{\rm{N/m}}\).

Taking equilibrium position of the spring as the reference position, the elastic potential energy of the spring when stretched by a distance \(x'\)from the equilibrium position is as follows:

\(\begin{array}{c}P{{E'}_{{\rm{el}}}} = \frac{1}{2}k{{x'}^2}\\ = \frac{1}{2}\left( {3.0 \times {{10}^4}\;{\rm{N/m}}} \right){\left( {6.0 \times {{10}^{ - 2}}\;{\rm{m}}} \right)^2}\\ = 54\;{\rm{J}}\end{array}\)

The potential energy of spring changes when external force does work to change its position. Therefore, work required to stretch the spring from distance x to\(x'\)will be equal to the change in potential energy of the spring in going from distance x to\(x'\), i.e.,

\(\begin{array}{c}W = P{{E'}_{{\rm{el}}}} - P{E_{{\rm{el}}}}\\ = \left( {54 - 6.0} \right)\;{\rm{J}}\\ = 48\;{\rm{J}}\end{array}\)

Thus, work required to stretch the spring an additional 4.0 cm is 48 J.