The Physics
Hypertextbook
Opus in profectus

# Conservation of Energy

## Practice

### practice problem 1

The diagram below shows a 10,000 kg bus traveling on a straight road which rises and falls. The horizontal dimension has been foreshortened. The speed of the bus at point A is 26.82 m/s (60 mph). The engine has been disengaged and the bus is coasting. Friction and air resistance are assumed negligible. The numbers on the left show the altitude above sea level in meters. The letters A–F correspond to points on the road at these altitudes.

Magnify

1. Find the speed of the bus at point B.
2. 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.
3. Derive an equation to determine the speed of the bus at any altitude.

### practice problem 2

Two 64 kg stick figures are performing an extreme blob jump as shown in the diagram below. (Warning: These are professional stunt stick figures. Don't try this at home.)

One stick figure stands atop a 7.0 m high platform with a 256 kg boulder. A second stick figure stands a partially inflated air bag known as a blob (or water trampoline). The first stick figure rolls the boulder off the edge of the platform. It falls onto the blob, catapulting the second stick figure into the air. What is the maximum height to which the second stick figure can rise? Assume that stick figures, boulders, and blobs obey the law of conservation of energy.

#### 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.

 Us = Ub msghs = mbghb mshs = mbhb (64 kg)hs = (256 kg)(7.0 m) hs = 28 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.

 mshs = mbhb ms4hb = 4mshb hs = 4hb = 4(7.0 m) hs = 28 m

### practice problem 3

The illustration below shows a vertical loop segment of a roller coaster. The path of the rails is highlighted in yellow.

Magnify

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.

Magnify

Summary of data given for this problem
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.

1. 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 ½g).
2. 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)?
3. 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

1. 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.

Fc = N + W

The problem said normal was half weight. We can do some simplification

 Fc = 12W + WFc = 32W

We have an equation for centripetal force and we have an equation for weight. Let's use them and then do some algebra.

 mva2 = 32mg ra va2 = 32g ra va = √32gra

 va = √[32(9.8 m/s2)(7.1 m)]va = 10.2 m/s
2. 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).

 Eb = Ea Kb + Ub = Ka + Ua ½mvb2 + mghb = ½mva2 + mgha

Algebra tells us to cancel out the common factor m (since it's in every term once) and get rid of the zero term mghb (since hb is zero).

½vb2 = ½va2 + gha

More algebra to isolate the desired quantity.

vb = √(va2 + 2gha)

Numbers go into the calculator.

vb = √((10.2 m/s)2 + 2(9.8 m/s2)(25.8 m))

Numbers come out of the calculator.

vb = 24.7 m/s

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.

Fc = 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 + Fc

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 gb. There are several ways to approach this problem. Here's my method…

gb′ =
 N W
gb′ =
 W + Fc W
gb′ =
 mg + mvb2/rb mg
gb′ =
 1 + vb2 grb

Cool bit of algebra. Let's finish this.

gb′ =
 1 + (24.7 m/s)2 (9.8 m/s2)(23.2 m)

gb′ = 3.68 g

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.

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

 Ec = Eb Kc + Uc = Kb + Ub ½mvc2 + mghc = ½mvb2 + mghb ½vc2 + ghc = ½vb2 + ghb ½vc2 + (9.8 m/s2)(13.4 m) = ½(24.7 m/s)2 + (9.8 m/s2)(0.0 m) vc = 18.6 m/s

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".)

gc′ =
 N W

gc′ =
 Fc W

gc′ =
 mvc2/rc mg

gc′ =
 vc2 grc

gc′ =
 (18.6 m/s)2 (9.8 m/s2)(13.4 m)

gc′ = 2.64 g

That's higher than normal in magnitude and unusual in direction. The table below summarizes the quantities computed for this problem.

Summary of 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

Write something completely different.