Question 26: (I) By how much does the gravitational potential energy of a 54-kg pole vaulter change if her center of mass rises about 4.0 m during the jump?

The energy possessed by the body by its position above the surface of the earth is termed gravitational potential energy.

For example, if an object of mass m is raised by a height h, the gravitational potential energy of the object at this height will be \(P{E_{\rm{G}}} = m{\rm{g}}h\).

Here, g is the acceleration due to gravity on the surface of the earth.

Mass of pole vaulter is m = 54 kg.

Centre of mass of pole vaulter rises by the height, h = 4.0 m.

If the ground is taken as the reference level, then the potential energy of the pole vaulter before the jump would be zero. Thus, the change in gravitational potential energy of the pole vaulter during the jump will be equal to his gravitational potential energy at height h. Therefore,

\(\begin{array}{c}\Delta P{E_{\rm{G}}} = m{\rm{g}}h\\ = \left( {54\;{\rm{kg}}} \right) \times \left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {4.0\;{\rm{m}}} \right)\\ = 2116.8\;{\rm{J}}\\ = 2.{\rm{1}} \times {\rm{1}}{{\rm{0}}^2}\;{\rm{J}}\end{array}\)

Thus, the change in gravitational potential energy of the pole vaulter is \(2.{\rm{1}} \times {\rm{1}}{{\rm{0}}^2}\;{\rm{J}}\).