Question 27: (I) A spring has a spring constant k of 88.0 N/m. How much must this spring be compressed to store 45.0 J of potential energy?

The energy associated with the state of compression or expansion of an elastic object is termed the elastic potential energy of that object.

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

The spring constant of spring is k = 88.0 N/m.

The elastic potential energy of the spring is \(P{E_{{\rm{el}}}} = 45.0\;{\rm{J}}\).

Let the spring be compressed by a distance x from its natural position.

If the natural position of the spring is taken as the reference position, the elastic potential energy of the spring at its natural position will be zero. Thus, potential energy stored in the spring on compressing it by a distance x is equal to the elastic potential energy of the spring,i.e.,

\(\begin{array}{c}P{E_{{\rm{el}}}} = \frac{1}{2}k{x^2}\\45.0\;{\rm{J}} = \frac{1}{2}\left( {88.0\;{\rm{N/m}}} \right){x^2}\\{x^2} = \frac{{2 \times 45.0\;{\rm{J}}}}{{88.0\;{\rm{N/m}}}}\\{x^2} = 1.023\;{{\rm{m}}^2}\\x = 1.01\;{\rm{m}}\end{array}\)

Thus, the spring is compressed by a distance of 1.01 m.