Conservation of Energy
Written by tutor Zhanna S.
This content is intended for the general physics without calculus.
Non-Isolated & Isolated systems, law of conservation of energy
To solve problems with work and energy one needs to identify a system that one is working with.
A non-isolated system is a system in which the energy can cross its boundary; however, the total amount of energy in the universe is conserved. An example of a non-isolated system can be a pot (with or without a cover) on the stove with boiling water. As water boils, the water vapor escapes the pot (through the crevices of the cover if there’s cover present). The energy crosses the boundary (the top of the pot) of the non-isolated system. The change in system’s total energy is changed, but the amount of energy in the universe is conserved since the water was converted to water vapor and heat, not lost into oblivion. The energy is neither created or destroyed, just converted into other form of energy such as thermal energy, radiation, etc. The human is a non-conservative (non-isolated) system. Human interacts with the environment. Human eats to get chemical energy, which is then converted into mechanical energy to walk to work and do work. More on the non-isolated system can be found in calculus-based physics textbooks (e.g. Serway, Jewett, Physics for Scientists and Engineers with Modern Physics, Eighth edition, p. 199-234).
The total energy of an isolated system can be represented/imagined with or without boundaries. Some examples of an isolated system can range from a small box (like a vacuumed chamber containing gas particles bouncing off the walls of the container, in this case sulfur dioxide) to a classroom, college building, book-Earth system, state, country, solar system, and even the whole universe.
Why do you think the coffee stays hot for so long in the closed-lid container? Not much energy is lost to the surroundings or to friction on the walls as you lift up the cup and drink the coffee. So the energy/temperature of this isolated system stays constant for some period of time.
This brings us to:
The law of conservation of energy:
the total amount of energy of an isolated system is conserved/constant.
Mathematically stated:
E_{final of a system} = E_{initial of a system} (1)
Conservative and nonconservative forces
To differentiate whether you are dealing with an isolated or a non-isolated system, consider the forces acting on the system.
Conservative forces acting on the system such as the force due to gravity, Fg, or the spring force, Fs only depend on initial and final points of the object. Hence,
a force is conservative if the work done by that force in moving an object is path-independent or if the work done by moving the object round –trip is equal to zero. Examples can include lifting up a can of soda and placing it back on the table. The work done is work opposing gravity and Force of gravity is the vertical force and is only dependent on height, independent of horizontal path. When you lift the can and place it back on the table the total work is zero since W_{1} = F*d, W_{2} = F*(-d).
A force is nonconservative if the work done by that force in moving an object is path-dependent. An example of a nonconcervative force is applied force, such as push, pull, and the force of friction, F_{f}. More work is done in moving a crate to the corner by going around the room as depicted below than just moving it across the room.
The force of friction was opposing the movement of the object 7 meters in the first case (more work done) and 5 meters in the second case (less work done).
Conservation of energy vs. conservation of total mechanical energy
The energy transfer in/out of a system can be accomplished by work, mechanical waves, radiation, heat transfer, etc. For an isolated system, the sum of all potential (U) and all kinetic (K) energies is constant and equals to the total mechanical energy (E). Mathematically represented as:
E_{mechanical} = ΣK + ΣU | (2) |
E_{initial} = E_{final} | (3) |
The initial and final energies of an isolated system are the kinetic and potential energies:
E_{i} = K_{i} + U_{i} | E_{f} = K_{f} + U_{f} | (4) |
K_{i} = ^{1}/_{2}mv_{i}^{2} | K_{f} = ^{1}/_{2}mv_{f}^{2} | (5) |
U_{i} = mgh_{i} | U_{f} = mgh_{f} | (6) |
Recall that potential energy of a spring depends on the position x, the spring either stretched or compressed from the equilibrium. Usually x_{i} is taken as 0.
U_{si} = ^{1}/_{2}kx_{i}^{2} U_{sf} = ^{1}/_{2}kx_{f}^{2} (7)
If we have more than 1 particle then the kinetic energies of all particles are added. Pay close attention to the subscripts. Differentiate initial and final states of the system for each particle.
K_{i} = ^{1}/_{2}m_{1}v_{1i}^{2} + ^{1}/_{2}m_{2}v_{2i}^{2} + … . | |
K_{f} = ^{1}/_{2}m_{1}v_{1f}^{2} + ^{1}/_{2}m_{2}v_{2f}^{2} + … . | (8) |
The same is done with the potential energy if we have more than 1 particle:
U_{i} = m_{1}gh_{1i} + m_{2}gh_{2i} + … . | |
U_{f} = m_{1}gh_{1f} + m_{2}gh_{2f} + … . | (8) |
Also, in a conservative system one form of energy is transferred into another and the sum of the changes is zero. Mathematically it is expressed as:
ΔK + ΔU = 0 (9)
When we expand equation 10 more detailed we remember that any change is the difference between the final and initial states:
ΔK = K_{final} – K_{initial} (10)
Let’s look at some examples.
Example 1: Thor stands on the top of the 92-story (423m) Stark building and fights Loki with his hammer. He is deceived by Loki and drops his hammer from the height of the building. Assuming the hammer is made from the star and its mass is 300 billion elephants, or 4.5 quadrillion pounds.
a) What is the potential energy of the hammer when it is in Thor’s hand?
b) What is the ratio of this energy to the energy of 1 Oreo cookie? Assume 1 cookie is 55 calories. How many cookies would you need to eat to get that much energy?
c) What is the kinetic energy of the hammer when it has fallen ^{1}/_{3} of the way down?
d) What is the final speed of the hammer when it strikes the ground?
e) How does this speed compare to the car traveling at 70mph?
Solution:
a) Potential energy of the hammer = ?
1. Think about the problem. Do you need to do any further research or make any assumptions? Usually you don’t, but here you do.
In this case do you assume that hammer falls from Thor’s hand or from the top of the building? Let’s say from his hand. Then you need to know how tall he is. Let’s assume that he is 2m tall. Then the total height the hammer falls is 425m.
2. Set up your problem. Write out all the givens and draw a picture.
Given:
Height initial = h_{i} = 425m
Mass = 4.5 * 1,000,000,000,000,000 lb = 4.5 * 10^{15} lb
a) U_{at the top} = U_{i} = ?
b) Ratio of energies = ?
c) K = ? ^{1}/_{3} way down
d) v_{f} = ?
3. Convert all units to the metric system. Use a search engine or reference book, if needed. In this case 1kg = ? lb. I find online that 1kg = 2.20lb
4.5 * 10^{15}lb * ^{1kg}/_{2.20 lb} = 2.05 * 10^{15} kg = mass of the hammer
4. Use the basic principles of conservation of energy. If we imagine our system with no friction present (no air resistance), then we have eq. 3:
E_{i} = E_{f}
Identifying equation 3 in terms of kinetic and potential energies we use equations 4, 5, 6.
E_{i} = K_{i} + U_{i} = ^{1}/_{2}mv_{i}^{2} + mgh_{i}
Substituting the numbers, noting that initial velocity of the hammer in Thor’s hand is zero since it did not start falling yet, we have:
E_{i} = ^{1}/_{2}(2.05 * 10^{15} kg)(0)^{2} + (2.05 * 10^{15} kg) (9.8 ^{m}/_{s2})(425 m)
E_{i} = 8.54 * 10^{18} ^{kg*m2}/_{s2} = 8.54 * 10^{18} J
So, for part A of the example question, the answer is 8.54 * 10^{18} J.
b) Ratio of energies = ?
Here, we have to look up conversion factor (using a reference book or search engine online) from calories to joules because the energy of the hammer is measured in joules and we can only make meaningful comparisons with the same units.
1 cal = 4.184 Joules
Energy of 1 cookie = 55 cal * ^{4.184J}/_{1 cal} = 230.12 J
Ratio = ^{Energy of hammer}/_{Energy of cookie} = ^{8.54 * 1018 J}/_{230.12 J} = 3.71 * 10^{16}
Therefore Thor’s hammer is that many times more powerful than the Oreo cookie.
To get the energy of a hammer we need to eat:
8.54 * 10^{18} J * ^{1 cookie}/_{230.12 J} = 3.71 * 10^{16} cookies!
So, for part B of the example question, the answer is that Thor’s hammer is 3.71 * 10^{16} times more powerful than the Oreo cookie.
c) K = ? ^{1}/_{3} of the way down
While the hammer is falling, the energy is still conserved and the initial potential energy is converted into kinetic energy. So we use:
E_{initial} = E_{final}
from part (a) we found that initial energy is 8.54 * 10^{18} J, which is also E_{f}. We don’t need initial energy. We need final kinetic energy, or energy ^{1}/_{3} of the way down.
From eq. 4 we have:
E_{f} = K_{f} + U_{f}
rearranging terms:
K_{f} = E_{f} – U_{f}
E_{f} = 8.54 * 10^{18} J,
U_{f} = mgh_{f}
What height is this equation referring to? The height when the hammer is ^{1}/_{3} of the way down is ^{1}/_{3}*425m =141.7m. That means the hammer fell 141.7 meters down, and is now at the height of 425m-141.7m = 283.3m
U_{f} = mgh_{f} = (2.05 * 10^{15} kg)(9.8 ^{m}/_{s2})(283m) = 5.69 * 10^{18} J
K_{f} = 8.54 * 10^{18} J – 5.69 * 10^{18} J = 2.85 * 10^{18} J
Therefore, the answer to part C of the example question is that the kinetic energy of the hammer ^{1}/_{3} of the way down is 2.85 * 10^{18} J.
v_{f} = ?
Before hammer hits the ground (h=0) all the potential energy (mgh) is converted into kinetic energy:
E_{f} = K_{f} + U_{f}
and from eq. 5 we have:
K_{f} = ^{1}/_{2}mv_{f}^{2}
Solving for v, we have:
2K_{f} = mv_{f}^{2}
^{2Kf}/_{m} = v_{f}^{2}
____ | _______ | ||||
v = | √ | v_{f}^{2} | = | √ | ^{2Kf}/_{m} |
__________ | _______ | ||||
v = | √ | ^{2(2.85*1018J)}/_{2.05*1015 kg} | = 52.73 | √ | ^{kg*m2}/_{s2} |
=52.73 ^{m}/_{s}
Therefore, the answer to part D of the example problem is that the final speed of the hammer is 52.73 ^{m}/_{s}.
e) How does this speed compare to the car traveling at 70mph?
To compare this speed to a car traveling at 70mph we need to have the same units. So we convert m/s to mi/hr:
^{52.73m}/_{s} * ^{60s}/_{1min} * ^{60min}/_{1 hr} * ^{1km}/_{1000m} * ^{1mi}/_{1.61km} = 117.906 mi/hr
Therefore, the answer to part E of the example problem is that the hammer was traveling at 117.906 mph while the car was going only 70mph. Therefore, the hammer was going almost twice as fast as the car.
Example 2: A 7.5-kg Sonic runs on ice horizontally. He runs into a spring with a speed of 157 ^{mi}/_{hr}, which projects him backwards. The spring constant is 8.46*10^{4} ^{N}/_{m}.
a) What is the total energy of the system?
b) What’s Sonic’s kinetic energy when the spring is compressed 33cm?
c) What is the total energy of the system if Sonic and Tails (the fox, 6.3-kg) run into the spring together?
m_{s} = 7.5kg
m_{t} = 6.3 kg
v_{i} = 157mph = 70.17m/s
(can make conversions right away)
k = 8.46*10^{4}N/m
x_{i} = 0
x_{f} = 33cm = 0.33m
a) E_{sys} = ?
Here, we are dealing with conservative forces again. We do not have friction because ice is slippery we assume that friction can be neglected. Therefore, we apply the principle of conservation of energy. Recall equation 7 for the potential energy of a spring. Initial position of the spring is x=0, so we only have kinetic energy right before Sonic runs into the spring.
E_{i} = E_{f}
E_{i} = K_{i} + U_{i} = ^{1}/_{2}mv_{i}^{2} + ^{1}/_{2}kx_{i}^{2} = ^{1}/_{2}(7.5kg)(70.17 ^{m}/_{s})^{2} = 18464.4 J
By conservation of energy no energy is lost in the system when sonic collides with the spring (so our assumption is that no heat is given off during collision), therefore the obtained value is the total energy of the system.
Therefore, the final answer for part A of example problem 2 is that there are 18464.4 J in the system.
b) K_{s} = ? at x_{f} = 0.33m
E_{f} = K_{f} + U_{f}
K_{f} = ?
K_{f} = U_{f} – E_{f}
since E_{i} = E_{f}
K_{f} = U_{f} – E_{i}
K_{f} = ^{1}/_{2}kx_{f}^{2} – 18464.4 J = ^{1}/_{2}(8.46 * 10^{4} ^{N}/_{m})(0.33m)^{2} = 4606.47^{kg*m/s2}/_{m} * m^{2}
K_{f} = 4606.47 J
Therefore, the answer to part B of the practice problem is that Sonic’s kinetic energy when the spring is compressed is 4606.47 J.
c) E_{(Sonic + Tails)}
Expanding part a), we have to account for 2 masses within the system. We know already that potential energy when spring is in equilibrium is zero, so the energy of a system is the kinetic energy of 2 bodies. Here, we have to assume that Tails runs with the same speed as Sonic to catch up to him, so his speed is the same as Sonic’s speed. So we expect energy almost to double.
E_{i} = K_{i s} + K_{i t} = ^{1}/_{2}mv_{i s}^{2} + ^{1}/_{2}mv_{i t}^{2} = ^{1}/_{2}(7.5 kg)(70.17 ^{m}/_{2})^{2} + ^{1}/_{2}(6.3kg)(70.17 ^{m}/_{s})^{2}
E_{i} = 33974.4 J
Therefore, the answer to part C of the example problem is that the total energy, if Sonic and Tails run into the spring together, is 33974.4 J.
Nonconservative forces
Recall, from the work-energy theorem, that the net work done on the system is the work done by conservative and nonconservative forces:
W_{c} + W_{nc} = ΔK + ΔU = ΔE (12)
Usually the work done by the non-conservative forces will result in dissipation of energy and the loss of energy so that
E_{i} ≠ E_{f} (13)
for a nonconservative system. Recall that the force of kinetic friction is one of the most common nonconservative forces,
F_{k}=μ_{k}N (14)
where μ_{k} is the coefficient of kinetic friction and differs from surface to surface. It will be given in the problem. Also recall that work equals the component of force parallel the object’s motion times distance the object moves:
W=F cosΘ*d (15)
Let’s look at some examples with the non-conservative forces.
Example 3: Captain America slides down the hill in the winter sitting on his shield. Iron Man measures his speed at the bottom of the hill to be 10^{m}/_{s}. The hill is 50m high and Captain America weighs 90kg.
a) Is the system conservative or nonconservative? Explain.
b) How much work was done by friction?
Given:
v_{i}=0
v_{f}=10^{m}/_{s}
Δh=50m
m=90kg
a) Does E_{i} = E_{f}?
b) W_{nc} = ?
a) Does E_{i} = E_{f}?
E_{i} = K_{i} + U_{i} = 0 + mgh_{i} = 90kg * (9.8 ^{m}/_{s2})(50m) = 44100 J
E_{f} = K_{f} + U_{f} = ^{1}/_{2}mv_{f}^{2} + 0 = ^{1}/_{2}(90kg)(10 ^{m}/_{s})^{2} = 4500 J
Therefore, the final answer for part A of example problem 3 is E_{i} ≠ E_{f}
Since initial and final energies are not equal the system is nonconservative.
b) W_{nc} = ?
From eq. 12 above:
W_{c} + W_{nc} = ΔK + ΔU = ΔE
W_{nc} = ΔE = E_{f} – E_{i} = 4500 J – 44100 J = -39600 J
Therefore, the final answer for part B of example problem 3 is that the work done by friction is -39600 J
Recall that the negative sign of work means it opposes the direction of motion, which is what we would expect from a frictional force.
Example 4: A 0.17 kg hockey puck slides on the ice with a speed of 85 ^{mi}/_{hr}. Suddenly, it goes over a rough patch of ice 70 cm in length and then back onto the smooth frictionless surface.
a) What is the final speed of the hockey puck after the rough patch if the coefficient of kinetic friction between the hockey puck and the rough patch is 0.93?
b) What would be the final velocity of an 84kg hockey player who fell down and slid on the rough patch with initial velocity of 30^{mi}/_{hr}?
Given:
m_{puck}= 0.17kg
m_{player}= 84kg
v_{i, puck} = 85^{mi}/_{hr} = 38.0 ^{m}/_{s}
v_{i, player}= 30^{mi}/_{hr} = 13.4 ^{m}/_{s}
x_{i} = 0
x_{f} = 0.70m
μ_{k}=0.93
a) v_{f, puck} = ?
b) v_{f, player} = ?
v_{f, puck} = ?
W_{nc} = ΔE = E_{f} – E_{i} = K_{f} – K_{i} since there is no potential energy (height = 0)
The work due to friction is the nonconservative work. So the left hand side of the equation becomes:
W_{nc} = F_{k} * d * cos(180°)
since force of kinetic friction and distance traveled are 180 degrees apart.
From eq. 14, we sub in for force of friction
W_{nc} = μ_{k}N * d * (-1)
Recalling that the normal force N is the opposite of gravity, or weight, mg, we rewrite
W_{nc} = -μ_{k}mgd
Now we plug in this result instead of nonconservative work in the first equation:
-μ_{k}mgd = K_{f} – K_{i}
-μ_{k}mgd = ^{1}/_{2}mv_{f}^{2} – ^{1}/_{2}mv_{i}^{2}
Solving for final velocity we divide by m, multiply by 2:
-2μ_{k}gd = v_{f}^{2} – v_{i}^{2}
-2μ_{k}gd + v_{i}^{2} = v_{f}^{2}
___ | __________ | |||||
v_{f} | = | √ | v_{f}^{2} | = | √ | -2μ_{k}gd + v_{i}^{2} |
_______________________________ | |||||
v_{f} | = | √ | -2(0.93)(9.8 ^{m}/_{s2})(0.70m) + (38.0 ^{m}/_{s})^{2} | = | 37.83^{ m}/_{s} |
b) v_{f, player} = ?
_______________________________ | |||||
v_{f} | = | √ | -2(0.93)(9.8 ^{m}/_{s2})(0.70m) + (13.4 ^{m}/_{s})^{2} | = | 12.92^{ m}/_{s} |
Note that the final velocity is independent of mass but dependent on coefficient of friction and initial velocity.
Conservation of Energy Practice Quiz
A child on a sled rides down from the top of the 20m hill. If his initial speed was 3m/s, how much speed did he gain by the time he got to the bottom of the hill?
Hint:
The gain of speed is the difference between initial and final speeds.
3 m/s
17 m/s
20 m/s
The child does not gain speed.
Δv = v_{f} = v_{i} = 17 m/s
Hulk throws 5 rocks at the invading-Earth aliens from the top of the skyscraper. He throws 5 rocks at the same time with just 1 hand and they go different directions. Which rock will hurt the most (greatest velocity)? See diagram below for the directions of the rocks:
Angles 1, 2, 3, and 4 are 60, 45, 50, and 70 degrees respectively as measured from the horizontal. All the rocks are the same size. Neglect air resistance.
The gain of speed is the difference between initial and final speeds.
1 & 2 will have the same velocity
2 & 4 will have the same velocity
3 will have the greatest velocity
They will all have the same velocity.
All rocks will hurt the same -have the same velocity. Energy is composed of Kinetic and Potential energies. In this case, all the rocks have the same initial kinetic energies (velocities, since thrown with the same hand). Potential energy is only dependent on the height and is independent of the horizontal path, hence, it’s independent of the angle.
The Hulk is standing on the ground. He throws a 30kg rock at a 90degree angle (vertically upward). If the initial speed is 20m/s, what is the c) total mechanical energy?
6000 J
0 J
147000 J
Not enough information.
E = K + U = 6000 J