- Work (W) done on an object by a force (F) over a distance (d):
W = F • d
- Kinetic energy:
KE = mv2/2
- (Gravitational) potential energy:
PE = mgh
- Total Energy (without friction):
Etot = KE + PE
Etot = mv2/2 + mgh
- Total Energy (with friction):
Etot = KE + PE + Wf
Etot = mv2/2 + mgh + Wf
(where Wf is the work done by friction)
P = Work/Time
P = W/t
A) If it takes a force of 2760 N [parallel to the track] to pull the roller coaster up to point A, and it is pulled along 150 m of track, find the work done on the roller coaster.
We need the work done on the roller coaster, which is the net work, or the work done by all the forces on the roller coaster. First, there is a 2760 N force on the coaster, which acts along a distance of 150 m. There is also the work done by gravity, which spans a distance of 65 m (up to point A).
First we find the work done by the force of 2760 N up a track along a distance of 150 m. The work done by this force is
W1 = F • d = (2760 N)(150 m) = 414,000 J = 414 kJ
This number should be positive since the force is adding to the total energy of the roller coaster. On the other hand, if a force removes energy from an object (like friction), then the work done would be negative.
Next, we find the work done by gravity. We need a force, so we need to find the weight of roller coaster, which is given by w = mg = (650 kg)(9.8 m/s2) = 6370 N. The force of gravity does work on the roller coaster only as it changes height, so the total distance that gravity does work on is the total height of our path, which is 65 m. We have to remember that as the object travels upwards, gravity works against the direction of motion, so the work done by gravity should be negative. There is a mathematical way to show this, but this requires a diagram and some more explanation (specifically, it requires that we use the equation W = F • d • cosθ where θ is the angle between the displacement vector and the force vector, and this will always give us the proper negative sign). For now, we have to make sure to insert a negative sign in our answer. The work done by gravity is then
W2 = F • d = - (6370 N)(65m) = - 414,050 J ≈ - 414 kJ
We've just calculated the work done by the first force, and then by gravity. The total work is just the sum of all the work:
Wtot = W1 + W2 = 414 kJ + (-414 kJ) = 414 kJ - 414 kJ = 0 J
Wtot = 0 J
B) Find the total energy the roller coaster has at point A if it is at rest for a brief moment before it starts to go down the hill.
This question asks for the total energy, which is given by Etot = KE + PE = mv2/2 + mgh. At point A, the roller coaster is at rest, so v = 0 m/s (or KE = 0). Therefore, the total energy reduces to
Etot = PE = mgh.
The height here is h = 65 m, so the total energy is
Etot = PE = mgh = (650 kg)(9.8 m/s2)(65 m) = 414,050 J ≈ 414 kJ.
C) According to the Law of Conservation of Energy, what will the total energy of the roller coaster be at points B and C?
The Law of Conservation of Energy simply states that, as an object changes heights or changes speeds (in general, as it goes through some sort of change), the total energy (Etot) of the object remains the same. In other words, total energy cannot be created or destroyed, but can simply change forms (like from kinetic energy to potential energy). When friction or engines are introduced, the total energy of the object may no longer be the same, but we should still be able to account for all the energy, even the energy dissipated or turned into heat by friction.
Mathematically, we can say that
EA = EB = EC
which is saying that the total energy at point A is the same at point B and the same at point C. The kinetic and potential energies may differ, but the total energy must always be same. Therefore,
EA = EB = EC = 414 kJ
D) Use the Law of Conservation of Energy to find the speed of the roller coaster at point B.
At point B, we found the total energy to be
Etot = KE + PE = mv2/2 + mgh (at point B)
or, since Etot = 414 kJ,
414 kJ = mv2/2 + mgh
We know the height at point B is h = 52 m, so we now have
414 kJ = mv2/2 + (650 kg)(9.8 m/s2)(52 m)
414 kJ = mv2/2 + 331.2 kJ
Insert the mass as m = 650 kg:
414 kJ = (650 kg)v2/2 + 331.2 kJ
414 kJ = 325•v2 + 331.2 kJ
Subtract 331.2 kJ from both sides:
82.8 kJ = 325•v2
Convert kJ to Joules:
82,800 J =325•v2
Divide both sides by 325:
254.8 = v2
Take the square root of both sides:
v = 15.96 m/s
E) If it takes the roller coaster 3 minutes to travel around the track, find the power of the roller coaster.
Power, measured in Joules per second (or Watts), is the total work done on an object per unit time, or
P = Work/Time
The power of the roller coaster is simply the work done by the roller coaster divided by the total amount of time taken. We calculated the work in part A, which was
W = F • d = (2760 N)(150 m) = 414,000 J = 414 kJ
(In part A, we called it W1, but here we'll just call it W).
The total time is 3 minutes, but we need the time in seconds. Since there are 60 seconds per minute, three minutes is simply
t = 3 min • (60 seconds/ 1 minute) = 180 s
The power is then
P = W/t = 414 kJ/180 s = 2.3 kW = 2,300 Watts
F) In real life, some of the energy the roller coaster has at point A will be lost as thermal energy due to friction as it travels through the track. If the speed of the roller coaster at point C is only 19 m/s, calculate the amount of energy that was lost due to friction.
At point C, we can still make the claim that the total energy is 414 kJ (from part C) as long as we include friction into the total energy calculation. In other words, we can say that the total energy is the sum of potential energy (PE), kinetic energy (KE), and work done by friction (Wf):
Etot = KE + PE + Wf
= mv2/2 + mgh + Wf
We know that the total energy is Etot = 414 kJ, the mass is m = 650 kg, the speed is v = 19 m/s, and the height is h = 25 m (given in the problem statement). Plugging these values in,
Etot = mv2/2 + mgh + Wf
414 kJ = (650 kg)(19 m/s)2/2 + (650 kg)(9.8 m/s2)(25 m) + Wf
414 kJ = 276.6 kJ + Wf
Subtracting 276.6 kJ from both sides gives
Wf = 137.4 kJ = 137,400 J