# 8.2: Potential Energy of a System

- Page ID
- 4013

- What is the difference in just the spring potential energy, from an initial equilibrium position to its pulled-down position?
- What is the difference in just the gravitational potential energy from its initial equilibrium position to its pulled-down position?
- What is the kinetic energy of the block as it passes through the equilibrium position from its pulled-down position?

**Strategy**

In parts (a) and (b), we want to find a difference in potential energy, so we can use Equations \ref{8.6} and \ref{8.4}, respectively. Each of these expressions takes into consideration the change in the energy relative to another position, further emphasizing that potential energy is calculated with a reference or second point in mind. By choosing the conventions of the lowest point in the diagram where the gravitational potential energy is zero and the equilibrium position of the spring where the elastic potential energy is zero, these differences in energies can now be calculated. In part (c), we take a look at the differences between the two potential energies. The difference between the two results in kinetic energy, since there is no friction or drag in this system that can take energy from the system.

**Solution**

- Since the gravitational potential energy is zero at the lowest point, the change in gravitational potential energy is $$\Delta U_{grav} = mgy - 0 = (12\; N)(5.0\; cm) = 0.60\; J \ldotp$$
- The equilibrium position of the spring is defined as zero potential energy. Therefore, the change in elastic potential energy is $$\Delta U_{elastic} = 0 - \frac{1}{2}ky_{pull}^{2} = - \left(\dfrac{1}{2}\right) (6.0\; N/m) (5.0\; cm)^{2} = -0.5\; J \ldotp$$
- The block started off being pulled downward with a relative potential energy of 0.75 J. The gravitational potential energy required to rise 5.0 cm is 0.60 J. The energy remaining at this equilibrium position must be kinetic energy. We can solve for this gain in kinetic energy from Equation 8.2, $$\Delta K = - (\Delta U_{elastic} + \Delta U_{grav}) = - (-0.75\; J + 0.60\; J) = 0.15\; J \ldotp$$

**Significance**

Even though the potential energies are relative to a chosen zero location, the solutions to this problem would be the same if the zero energy points were chosen at different locations.

Exercise \(\PageIndex{4}\)

Suppose the mass in Example 8.4 is in equilibrium, and you pull it down another 3.0 cm, making the pulled-down distance a total of 8.0 cm. The elastic potential energy of the spring increases, because you’re stretching it more, but the gravitational potential energy of the mass decreases, because you’re lowering it. Does the total potential energy increase, decrease, or remain the same?

Simulation

View this simulation to learn about conservation of energy with a skater! Build tracks, ramps and jumps for the skater and view the kinetic energy, potential energy and friction as he moves. You can also take the skater to different planets or even space!

A sample chart of a variety of energies is shown in Table \(\PageIndex{2}\) to give you an idea about typical energy values associated with certain events. Some of these are calculated using kinetic energy, whereas others are calculated by using quantities found in a form of potential energy that may not have been discussed at this point.

Object/phenomenon | Energy in joules |
---|---|

Big Bang | 10^{68} |

Annual world energy use | 4.0 x 10^{20} |

Large fusion bomb (9 megaton) | 3.8 x 10^{16} |

Hiroshima-size fission bomb (10 kiloton) | 4.2 x 10^{13} |

1 barrel crude oil | 5.9 x 10^{9} |

1 ton TNT | 4.2 x 10^{9} |

1 gallon of gasoline | 1.2 x 10^{8} |

Daily adult food intake (recommended) | 1.2 x 10^{7} |

1000-kg car at 90 km/h | 3.1 x 10^{5} |

Tennis ball at 100 km/h | 22 |

Mosquito (10^{−2} g at 0.5 m/s) | 1.3 x 10^{-6} |

Single electron in a TV tube beam | 4.0 x 10^{-15} |

Energy to break one DNA strand | 10^{-19} |

# Contributors

Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).