Question

Question 28: (II) If it requires 6.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additional 4.0 cm?

Short answer

The amount of work required to stretch the spring an additional 4.0 cm is 48 J.

Step-by-step solution

Elastic potential energy

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 \(P{E_{{\rm{el}}}} = \frac{1}{2}k{x^2}\).

Given information

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.

Determining spring constant 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 \(3.0 \times {10^4}\;{\rm{N/m}}\).

Determining new elastic potential energy of the spring when it is stretched 4.0 cm

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:

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

Determining work required to stretch the spring from a distance x to \(x'\) 

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.