# Difference between revisions of "Skill Set 06.1"

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− | [[IB Physics | + | [[Main Page]] > [[IB Physics Skill Sets]] > Skill Set 06.1 |

− | ==1. A 5.2 kg object speeds up from 3.1 m/s to 4.2 m/s. What is the change in kinetic energy? (21 J)== | + | ===1. A 5.2 kg object speeds up from 3.1 m/s to 4.2 m/s. What is the change in kinetic energy? (21 J)=== |

<blockquote> | <blockquote> | ||

This question is just to see if you can apply the basic formulas for energy, and solve for things like velocities and height. The basic IB formulas for this sort of thing are: | This question is just to see if you can apply the basic formulas for energy, and solve for things like velocities and height. The basic IB formulas for this sort of thing are: | ||

− | *W = Fs | + | *W = Fs<math> \cos \Theta </math> |

*<math>\Delta</math>E<sub>p</sub> = mg<math>\Delta</math>h | *<math>\Delta</math>E<sub>p</sub> = mg<math>\Delta</math>h | ||

*E<sub>k</sub> = <sup>1</sup>/<sub>2</sub>mv<sup>2</sup> | *E<sub>k</sub> = <sup>1</sup>/<sub>2</sub>mv<sup>2</sup> | ||

Line 12: | Line 12: | ||

*E<sub>p</sub> = <sup>1</sup>/<sub>2</sub>kx<sup>2</sup> | *E<sub>p</sub> = <sup>1</sup>/<sub>2</sub>kx<sup>2</sup> | ||

+ | |||

+ | So what you need to do is have your final kinetic energy minus the initial kinetic energy. <br/> | ||

+ | |||

+ | *E<sub>kf</sub> - E<sub>ki</sub> | ||

+ | <br/> <br/> | ||

+ | Your Final kinetic energy is just <math> \begin{matrix} \frac{1}{2} \end{matrix}</math> (5.2 kg)(4.2 m/s)<math> ^2 </math> = 45.864 J <br/> | ||

+ | And the initial energy is <math> \begin{matrix} \frac{1}{2} \end{matrix}</math>(5.2 kg)(3.1 m/s)<math>^2</math> = 24.986 J | ||

+ | <br/> <br/> | ||

+ | Then subtract<br/> | ||

+ | 45.864 J -24.986 J <br/><br/> | ||

+ | The answer is 20.878 J, but that has bad sig figs. <br/><br/> | ||

+ | So...<br/> | ||

+ | ROUND!<br> | ||

+ | <math> \Delta </math>E<sub>k</sub>=21 J | ||

+ | |||

+ | |||

</blockquote> | </blockquote> | ||

[[#top | Table of Contents]] | [[#top | Table of Contents]] | ||

− | ==2. A 1.2 HP motor (1 HP = 745.7 Watts) is used to raise a 1300 kg Land Rover 5.7 m up into a tree. What time will it take? (81 s)== | + | ===2. A 1.2 HP motor (1 HP = 745.7 Watts) is used to raise a 1300 kg Land Rover 5.7 m up into a tree. What time will it take? (81 s)=== |

+ | <blockquote> | ||

+ | This problem always has to do with calculating power. The data packet gives us | ||

+ | |||

+ | Power = Fv | ||

+ | |||

+ | Which is fairly useless. In general, Power is the rate at which work is done so Power is better explained as: | ||

+ | |||

+ | Power = <math>\frac{Any\ change\ in\ energy}{\Delta t}</math> | ||

+ | *W = Fs<math> \cos \Theta </math> | ||

+ | *<math>\Delta</math>E<sub>p</sub> = mg<math>\Delta</math>h | ||

+ | *E<sub>k</sub> = <sup>1</sup>/<sub>2</sub>mv<sup>2</sup> | ||

+ | Missing from the packet is | ||

+ | *E<sub>p</sub> = <sup>1</sup>/<sub>2</sub>kx<sup>2</sup> | ||

+ | </blockquote> | ||

+ | [[#top | Table of Contents]] | ||

− | ==3. A massless spring with a spring constant of 34 N/m is compressed 5.8 cm horizontally and used to shoot an 18 gram marble across a frictionless table. What is the speed of the marble? (2.5 m/s)== | + | ===3. A massless spring with a spring constant of 34 N/m is compressed 5.8 cm horizontally and used to shoot an 18 gram marble across a frictionless table. What is the speed of the marble? (2.5 m/s)=== |

− | ==4. A 3.4 kg bowling ball hanging from the ceiling on a long string swings from side to side like a pendulum. When it is 15 cm above its lowest point on the left side, I shove it with a force of 11 N for a distance of .35 m in the direction it is going. How high will it swing on the other side? (Neglect friction) (27 cm)== | + | ===4. A 3.4 kg bowling ball hanging from the ceiling on a long string swings from side to side like a pendulum. When it is 15 cm above its lowest point on the left side, I shove it with a force of 11 N for a distance of .35 m in the direction it is going. How high will it swing on the other side? (Neglect friction) (27 cm)=== |

− | ==5. A 580 kg rollercoaster is going 7.5 m/s on the top of a 1.2 m tall hill, how fast is it going on top of a 3.5 m tall hill? (Neglect friction) (3.3 m/s)== | + | ===5. A 580 kg rollercoaster is going 7.5 m/s on the top of a 1.2 m tall hill, how fast is it going on top of a 3.5 m tall hill? (Neglect friction) (3.3 m/s)=== |

## Latest revision as of 12:30, 3 March 2009

Main Page > IB Physics Skill Sets > Skill Set 06.1

## Contents

- 1 1. A 5.2 kg object speeds up from 3.1 m/s to 4.2 m/s. What is the change in kinetic energy? (21 J)
- 2 2. A 1.2 HP motor (1 HP = 745.7 Watts) is used to raise a 1300 kg Land Rover 5.7 m up into a tree. What time will it take? (81 s)
- 3 3. A massless spring with a spring constant of 34 N/m is compressed 5.8 cm horizontally and used to shoot an 18 gram marble across a frictionless table. What is the speed of the marble? (2.5 m/s)
- 4 4. A 3.4 kg bowling ball hanging from the ceiling on a long string swings from side to side like a pendulum. When it is 15 cm above its lowest point on the left side, I shove it with a force of 11 N for a distance of .35 m in the direction it is going. How high will it swing on the other side? (Neglect friction) (27 cm)
- 5 5. A 580 kg rollercoaster is going 7.5 m/s on the top of a 1.2 m tall hill, how fast is it going on top of a 3.5 m tall hill? (Neglect friction) (3.3 m/s)

### 1. A 5.2 kg object speeds up from 3.1 m/s to 4.2 m/s. What is the change in kinetic energy? (21 J)

This question is just to see if you can apply the basic formulas for energy, and solve for things like velocities and height. The basic IB formulas for this sort of thing are:

- W = Fs
- E
_{p}= mgh- E
_{k}=^{1}/_{2}mv^{2}Missing from the packet is

- E
_{p}=^{1}/_{2}kx^{2}So what you need to do is have your final kinetic energy minus the initial kinetic energy.

- E
_{kf}- E_{ki}

Your Final kinetic energy is just (5.2 kg)(4.2 m/s) = 45.864 J

And the initial energy is (5.2 kg)(3.1 m/s) = 24.986 J

Then subtract

45.864 J -24.986 J

The answer is 20.878 J, but that has bad sig figs.

So...

ROUND!

E_{k}=21 J

### 2. A 1.2 HP motor (1 HP = 745.7 Watts) is used to raise a 1300 kg Land Rover 5.7 m up into a tree. What time will it take? (81 s)

This problem always has to do with calculating power. The data packet gives us

Power = Fv

Which is fairly useless. In general, Power is the rate at which work is done so Power is better explained as:

Power =

- W = Fs
- E
_{p}= mgh- E
_{k}=^{1}/_{2}mv^{2}Missing from the packet is

- E
_{p}=^{1}/_{2}kx^{2}