# Conservation of Energy

## Practice

### practice problem 1

- Complete the table below. Let potential energy be zero at an altitude of 50 m.
A B C D E F speed 26.82 m/s height 55 m 50 m 55 m 60 m 65 m 60 m kinetic energy potential energy 0.000 MJ total energy - An extortionist has planted a bomb on the bus. If the speed of the bus falls below 22.35 m/s (50 mph) the bomb will explode. Will the speed of the bus fall below this value and explode? If you feel the bus will explode, identify the interval in which this occurs.

#### solution

Start with location A. All the needed information is given.

Use the kinetic energy equation.

*K*=_{A}½ *mv*_{A}^{2}*K*=_{A}½(10,000 kg)(26.82 m/s) ^{2}*K*=_{A}3,597,000 J Use the gravitational potential energy equation. Make the height relative to some reference point. I've decided to use the lowest point in the diagram — 50 m. At 55 m, location A is 5 m higher than that. You could choose any reference height you like. This was my choice.

∆ *U*=_{A}*mg*Δ*h*_{A}∆ *U*=_{A}(10,000 kg)(9.8 m/s ^{2})(55 m − 50 m)∆ *U*=_{A}490,000 J Energy being a scalar quantity, total energy is the simple sum of the kinetic and potential energies at this location.

*E*=_{A}*K*_{A}+*U*_{A}*E*=_{A}3,597,000 J + 490,000 J *E*=_{A}4,087,000 J Numbers this big won't fit in our data table unless we add a prefix to them. It looks like mega (M = 10

^{6}) has the right size for this problem. Time to complete the first column.A B C D E F speed 26.82 m/s height 55 m 50 m 55 m 60 m 65 m 60 m kinetic energy 3.597 MJ potential energy 0.490 MJ 0.000 MJ total energy 4.087 MJ With no friction or air resistance and no push coming from the engine, no energy is being subtracted or added to the bus. The total energy will stay the same. Let's complete that row of the table next.

A B C D E F speed 26.82 m/s height 55 m 50 m 55 m 60 m 65 m 60 m kinetic energy 3.597 MJ potential energy 0.490 MJ 0.000 MJ total energy 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ Location C is at the same height as location A. This means the bus has the same potential energy here that it did when it was at location A. The same total energy and the same potential energy implies the same kinetic energy. The same kinetic energy implies the same speed. Basically, just fill in the gaps in column C with values copied from column A.

A B C D E F speed 26.82 m/s 26.82 m/s height 55 m 50 m 55 m 60 m 65 m 60 m kinetic energy 3.597 MJ 3.597 MJ potential energy 0.490 MJ 0.000 MJ 0.490 MJ total energy 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ For three of the four remaining locations, determine the potential energy first. The potential energy at location B is trivial. It's zero, since that's what we've decided (or what I decided for you). Locations D and E require real calculation.

∆ *U*=_{B}*mg*Δ*h*_{B}∆ *U*=_{B}(10,000 kg) (9.8 m/s ^{2}) (50 m − 50 m)∆ *U*=_{B}0.000 MJ ∆ *U*=_{D}*mg*Δ*h*_{D}∆ *U*=_{D}(10,000 kg) (9.8 m/s ^{2}) (60 m − 50 m)∆ *U*=_{D}0.980 MJ ∆ *U*=_{E}*mg*Δ*h*_{E}∆ *U*=_{E}(10,000 kg) (9.8 m/s ^{2}) (65 m − 50 m)∆ *U*=_{E}1.470 MJ Kinetic energy is the difference between total energy and potential energy. Again, the solution for location B is trivial and the others require calculation.

*K*=_{B}*E*−_{B}*U*_{B}*K*=_{B}4.087 MJ − 0.000 MJ *K*=_{B}4.087 MJ *K*=_{D}*E*−_{D}*U*_{D}*K*=_{D}4.087 MJ − 0.980 MJ *K*=_{D}3.107 MJ *K*=_{E}*E*−_{E}*U*_{E}*K*=_{E}4.087 MJ − 1.470 MJ *K*=_{E}2.617 MJ Take the kinetic energy equation and rearrange it to make speed the subject of the equation.

*v*= √2 *K*⇐ *K*= ½*mv*^{2}*m*Use it over and over again, three times. Better change the energies from megajoules back into joules. Otherwise you'll break the equation.

*v*= √_{B}2 *K*_{B}*m**v*= √_{B}2(4,087,000 J) 10,000 kg *v*=_{B}28.59 m/s *v*= √_{D}2 *K*_{D}*m**v*= √_{D}2(3,107,000 J) 10,000 kg *v*=_{D}24.93 m/s *v*= √_{E}2 *K*_{E}*m**v*= √_{E}2(2,617,000 J) 10,000 kg *v*=_{E}22.88 m/s Now meet your new friend, Phillip D. Table.

A B C D E F speed 26.82 m/s 28.58 m/s 26.82 m/s 24.93 m/s 22.88 m/s height 55 m 50 m 55 m 60 m 65 m 60 m kinetic energy 3.597 MJ 4.087 MJ 3.597 MJ 3.107 MJ 2.617 MJ potential energy 0.490 MJ 0.000 MJ 0.490 MJ 0.980 MJ 1.407 MJ total energy 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ We still have location F left. It's at the same height as location D. No energy in or out of the system means those two locations have identical energies and speeds. Copy. Paste. Done.

A B C D E F speed 26.82 m/s 28.58 m/s 26.82 m/s 24.93 m/s 22.88 m/s 24.93 m/s height 55 m 50 m 55 m 60 m 65 m 60 m kinetic energy 3.597 MJ 4.087 MJ 3.597 MJ 3.107 MJ 2.617 MJ 3.107 MJ potential energy 0.490 MJ 0.000 MJ 0.490 MJ 0.980 MJ 1.407 MJ 0.980 MJ total energy 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ 4.087 MJ The bus will never move slower than it does at the highest point on the road — location E, where the speed is 22.88 m/s. Since this is still greater than 22.35 m/s (50 mph) the bus will not explode. The city bus will then be replaced with a cruise ship in a sequel that no one will watch.

### practice problem 2

#### solution

The situation starts with the boulder's gravitational potential energy (measured relative to the surface of the blob). The boulder falls and it's potential energy is transformed into kinetic energy. That kinetic energy gets transfered to the stick figure. Up goes the stick person. Kinetic energy is now transformed into potential energy. The energies at these four prominant times are all equal. Assuming energy was not lost, the initial potential energy of the boulder is equal to the final potential energy of the stick figure.

U = _{s} |
U_{b} |

m = _{s}gh_{s} |
m_{b}gh_{b} |

m = _{s}h_{s} |
m_{b}h_{b} |

(64 kg)h = _{s} |
(256 kg)(7.0 m) |

Another way to look at this problem is as a proportion. Potential energy is partly the product of mass and height. (It's also the product of gravity with mass and height, but since gravity doesn't change appreciably during a blob jump we can treat it as constant.) When the product of two numbers is contant, they are inversely proportional. The boulder has 4 times the mass of the stick figure. Therefore, the stick figure should have 4 times the height of the boulder.

m = _{s}h_{s} |
m_{b}h_{b} |

m4_{s}h = _{b} |
4m_{s}h_{b} |

h = _{s} |
4h = 4(7.0 m)_{b} |

### practice problem 3

Segments of roller coaster track are rarely circular. The transition between a straight segment and a circular segment, or two circular segments of different radii, would subject the rider to abrupt changes in acceleration, called jerk, that would be uncomfortable, especially at high speeds. Thrill rides should be thrilling, not jarring, jerking, or jostling. Curves with a gradually changing radius of curvature are more common.

The illustration below shows the same vertical loop with circles added to represent the instantaneous curvature at three locations. (For those of you who like technical language, these are called osculating circles.) The illustration also includes the radius of curvature and height above the lowest point on the track for these locations.

segment | h (m) |
r (m) |
---|---|---|

a. top of loop | 25.8 | 07.1 |

b. bottom of approach | 00.0 | 23.2 |

c. vertical descending side | 14.0 | 13.4 |

Assume minimal energy losses due to air resistance, rolling resistance, or other forms of friction and answer the following questions.

- Determine the speed of the coaster at the top of the loop if the normal force of the rails on the wheels is half the weight of the coaster (that is, if the frame of reference acceleration is ½ g).
- Determine the speed of the coaster at its lowest point before it entered the loop. How does the normal force on the wheels compare to the weight of the coaster now (that is, what is the new frame of reference acceleration)?
- Determine the speed of the coaster on the side of the loop when it is instantaneously moving straight down. How does the normal force on the wheels compare to the weight of the coaster at this location (that is, what is the frame of reference acceleration here)?

#### solution

This part of the problem is a circular motion problem and has nothing to do with conservation of energy yet. At the top of the loop, when the coaster is upside down, both weight at normal force point down. Together these forces provide the centripetal acceleration needed to make the turn. The problem said to ignore friction, so there are no forces acting left or right.

*F*=_{c}*N*+*W*The problem said normal was half weight. We can do some simplification

*F*=_{c}^{1}_{2}*W*+*W**F*=_{c}^{3}_{2}*W*We have an equation for centripetal force and we have an equation for weight. Let's use them and then do some algebra.

*mv*_{a}^{2}*r*_{a}= ^{3}_{2}*mg**v*_{a}^{2}*r*_{a}= ^{3}_{2}*g**v*_{a}= √ ^{3}_{2}*gr*_{a}We're ready for numbers.

*v*= √[_{a}^{3}_{2}(9.8 m/s^{2})(7.1 m)]*v*= 10.2 m/s_{a}Now we come to the conservation of energy part. With no friction or other outside forces, energy will be obviously conserved. The total mechanical energy at point b (where the coaster enters the loop) will equal the total mechanical energy at point a (the highest point in the loop).

*E*_{b}= *E*_{a}*K*+_{b}*U*_{b}= *K*+_{a}*U*_{a}½ *mv*_{b}^{2}+*mgh*_{b}= ½ *mv*_{a}^{2}+*mgh*_{a}Algebra tells us to cancel out the common factor

*m*(since it's in every term once) and get rid of the zero term*mgh*(since_{b}*h*is zero)._{b}½

*v*_{b}^{2}= ½*v*_{a}^{2}+*gh*_{a}More algebra to isolate the desired quantity.

*v*= √(_{b}*v*_{a}^{2}+ 2*gh*)_{a}Numbers go into the calculator.

*v*= √((10.2 m/s)_{b}^{2}+ 2(9.8 m/s^{2})(25.8 m))Numbers come out of the calculator.

*v*= 24.7 m/s_{b}The answer makes sense. The coaster travels fastest at the bottom and then slows as it works against gravity on its way to the top.

Now for the second half of the question. At the bottom of a vertical loop, weight points down like it

*always*does and normal points up like it*usually*does. Subtract these to get the centripetal force (the net force causing circular motion). Normal is the force pointing toward the center, so it should get the positive sign.*F*=_{c}*N*−*W*In other words, at the bottom of a curve, normal is greater than weight by an amount equal to the centripetal acceleration.

*N*=*W*+*F*_{c}Now let's figure out the number of gs we're pulling. How many times larger is the normal force on the coaster than its weight? I'll call that

*g*′. There are several ways to approach this problem. Here's my method…_{b}*g*′ =_{b}*N**W**g*′ =_{b}*W*+*F*_{c}*W**g*′ =_{b}*mg*+*mv*_{b}^{2}/*r*_{b}*mg**g*′ =_{b}1 + *v*_{b}^{2}*gr*_{b}Cool bit of algebra. Let's finish this.

*g*′ =_{b}1 + (24.7 m/s) ^{2}(9.8 m/s ^{2})(23.2 m)*g*′ = 3.68 g_{b}At the bottom of the loop, the coaster and it's passengers feel about 3½ times heavier than usual. This is both reasonably thrilling and reasonably safe. Remember, large accelerations can injure, maim, or even kill.

Back to the conservation of energy. Compare the energy at any two convenient places — say c and b. Substitute values and solve.

*E*_{c}= *E*_{b}*K*+_{c}*U*_{c}= *K*+_{b}*U*_{b}½ *mv*_{c}^{2}+*mgh*_{c}= ½ *mv*_{b}^{2}+*mgh*_{b}½ *v*_{c}^{2}+*gh*_{c}= ½ *v*_{b}^{2}+*gh*_{b}½ *v*_{c}^{2}+ (9.8 m/s^{2})(13.4 m)= ½(24.7 m/s) ^{2}+ (9.8 m/s^{2})(0.0 m)This answer makes sense. It's midway between the slowest speed at the top and the fastest speed at the bottom.

Continuing with the second half of the problem. At the side of the loop, weight still points down but normal now points to the right — in towards the center of curvature. Normal is the centripetal force and weight is counteracted by nothing. (Remember, there is no friction.) This means the coaster is sort of half in free fall. With nothing to balance weight, the coaster and its passengers are weightless in the vertical direction. There still is a normal force, so there still is an apparent gravity, but it points sideways!

Comparing normal to weight is one way to find the frame of reference acceleration. (It's also the first way the question was stated.) Let's try this and see what happens. (Be aware that the subscript c sometimes means "centripetal" and sometimes means "location c".)

*g*′ =_{c}*N**W**g*′ =_{c}*F*_{c}*W**g*′ =_{c}*mv*_{c}^{2}/*r*_{c}*mg**g*′ =_{c}*v*_{c}^{2}*gr*_{c}*g*′ =_{c}(18.6 m/s) ^{2}(9.8 m/s ^{2})(13.4 m)*g*′ = 2.64 g_{c}That's higher than normal in magnitude and unusual in direction. The table below summarizes the quantities computed for this problem.

segment | v |
g' |
|||
---|---|---|---|---|---|

a. | top of loop | 10.2 m/s | left | 0.50 g | down |

b. | bottom of approach | 24.7 m/s | right | 3.28 g | up |

c. | vertical descending side | 18.6 m/s | down | 2.64 g | right |

### practice problem 4

#### solution

Answer it.