# Springs

## Practice

### practice problem 1

mass (kg) |
position (cm) |
---|---|

0 | 026.5 |

1 | 040.0 |

2 | 054.0 |

3 | 068.5 |

5 | 097.5 |

6 | 112.5 |

7 | 127.5 |

8 | 143.5 |

#### solution

Hooke's law states that the force (*F*) on a spring is proportional to its extension (∆*x*).

*F* = *k*∆*x*

The proportionality constant in Hooke's law is called the spring constant (*k*).

k = |
F |

∆x |

The data given in this experiment were mass and position, not force and extension. The force in this experiment comes from the weight of the suspended masses. We have an equation for weight.

*W* = *mg*

Extension means change in length from some relaxed, natural, or unstressed length. Subtracting out the original position will give us the extension.

∆*x* = *x* − *x*_{0}

Compute these two quantities and change the extension from centimeters to meters (since that's the preferred SI unit). Compile the results in a table like the one below.

mass (kg) |
position (cm) |
force (N) |
extension (m) |
---|---|---|---|

0 | 026.5 | 00.0 | 0.000 |

1 | 040.0 | 09.8 | 0.135 |

2 | 054.0 | 19.6 | 0.275 |

3 | 068.5 | 29.4 | 0.420 |

5 | 097.5 | 49.0 | 0.710 |

6 | 112.5 | 58.8 | 0.860 |

7 | 127.5 | 68.6 | 1.010 |

8 | 143.5 | 78.4 | 1.170 |

Now graph it. Put force on the *y*-axis and extension on the *x*-axis. Since slope is…

m = |
∆y |

∆x |

…the slope of a line of best fit will give us the spring constant.

This spring from a chest expander has a spring constant of…

*k* = 67 N/m

### practice problem 2

mass of toy |
uncompressed height |
compressed height |
maximum height |
---|---|---|---|

16 g 0.016 kg |
11cm 0.11 m |
7 cm 0.07 m |
92 cm 0.92 m |

- the gravitational potential energy of the toy at its highest point relative to the table top
- the kinetic energy of the toy as it left the table
- the speed of the toy as it left the table
- the elastic potential energy of the toy after it was compressed
- the spring constant of the spring inside the toy

#### solution

Gravitational potential energy is found with the basic equation.

∆ *U*=_{g}*mg*∆*h*

∆*U*= (0.016 kg)(9.8 m/s_{g}^{2})(0.92 m)

∆*U*= 0.144 J_{g}The kinetic energy of the toy as it left the table is the same as the potential energy it had at the top of its flight. Energy changed form, but was conserved. There is nothing to calculate here.

*K*= 0.144 JUse the kinetic energy to determine the speed of the toy as it left the table.

*v*= √2 *K*⇐ *K*= ½*mv*^{2}*m*Be sure to use SI units.

*v*= √2 *K**m**v*= √2(0.144 J) (0.16 kg) *v*= 4.24 m/sThe elastic potential energy of the compressed toy is the same as the kinetic energy it had when it took off. Once again, energy was conserved and there's nothing to calculate.

*U*= 0.144 J_{s}Use the elastic potential energy equation to find the spring constant.

*k*=2 *U*_{s}⇐ *U*= ½_{s}*kx*^{2}*x*^{2}Watch out for non SI units. Don't let them interfere with a proper solution.

*k*=2 *U*_{s}*x*^{2}*k*= √2(0.144 J) (0.04 m) ^{2}*k*= 180 N/m

### practice problem 3

#### solution

Answer it.

### practice problem 4

#### solution

Answer it.