You can use Law of Conservation of Energy to solve physics problems. The law is that energy is not created or destroyed, so the total amount of energy stays the same, even though energy can be transferred or converted to other forms.
For a isolated system (no energy entering or leaving the system), you can use something like this:
Total Energy Before = Total Energy After
That might also be written like this:
KE + PE = KE’ + PE’
(KE’ is “KE prime” meaning KE at a later time)
If energy is entering or leaving the system, you add that into your equation. Energy that’s entering the system would be added on the left side, and energy leaving the system would add to the total on the right side.
Example 1
A 25 kg rock, with zero velocity, is 35 m above the ground. It is dropped and falls 35 m to the ground. How fast is it going just before it hits the ground?
Step 1: Write a conservation of energy equation.
In this example, the rock starts with gravitational potential energy and no kinetic energy. Just before it hits the ground, all of its energy has been converted to kinetic energy. Its gravitational potential energy at the beginning equals its kinetic energy at the end:
Eg = Ek
If you make a bar graph showing the types of energy involved, you write a term for each bar on each side of the equals signs.
(Explanation of bar chart illustrations on physicsclassroom.com)
We will now substitute the equations for gravitational energy and kinetic energy into this conservation of energy equation.
Reminder from Energy Notes:
Eg = mgh
Ek = ½mv2
Substituted Equation:
mgh = ½mv2
Now we can put all of the known values into the equation.
m = 25 kg
g = 10 m/s2 (Earth’s gravitational acceleration)
h = 35 m
v = ???
(25 kg)(10 m/s2)(35 m) = ½(25 kg)v2
8750 = (12.5)v2
700 = v2
26.5 m/s = v
(Note on units: the answer’s units really do come out to m/s, because all the kg units cancel out, and the m2/s2 units turn into m/s when you take the square root of both sides)
Example 2
In this example, there will be more than one type of energy on one of the sides of the equation, so it’s a little more complicated than example 1.
A 1200 kg car is coasting down a hill at 5 m/s. No friction is involved. It speeds up as it coasts down a vertical distance of 3.4 m. How much fast is it going now?
Conservation of Energy Equation:
Ek + Eg = Ek’
At the beginning of this situation, the car has both kinetic energy and gravitational potential energy, because it has some height (it’s higher up the hill than it will be at the end) and some velocity. At the end, after it has traveled down the hill a while, we’ll say that its height equals zero, and therefore its gravitational potential energy also equals zero. That’s why there is no Eg’ term on the right side. We could use any two height numbers we want, as long as the beginning height is 3.4 meters greater than the ending height. It’s easiest to call the starting height 3.4 and the final height zero.
Substituting Energy Equations into Conservation Equation:
½mv2 + mgh = ½mv’2
Note that there are two different velocities, v and v’, because the starting velocity is different than the final velocity)
Substitute numbers into equation
m = 1200 kg
v = 5.2 m/s
v’ = ???
g = 10 m/s2
h = 3.4 m
½(1200)(5.2)2 + (1200)(10)(3.4) = ½(1200)v’2
(I’ll multiply all of the numbers out next, but if you want to take a shortcut, you can cancel out all of the 1200 kg masses since it’s in every term in this equation)
16224 J + 40800 J = (600 kg)v’2
57024 J = (600 kg)v’2
(units note: J / kg = (m/s)2 so the units will work out properly)
95.04 = v’2
9.75 m/s = v’
Example 3
In this next example, I’ll add spring energy as one of the terms.
A motionless 0.80 kg block is being held against a spring at zero height. The spring is compressed so that it’s 0.44 m shorter than its resting length, and it has a spring constant of 1100 J/m2. After the block is released, it flies into the air. At a certain instant as the block is flying upward, its velocity is 13.1 m/s. How high is the block at that instant?
Conservation of Energy Equation
Es = Ek + Eg
At the start, there is only spring energy. The block is not moving and its position is at zero height. At the final instant, the spring has no energy left (its energy is transferred to the block when it springs), but it has velocity and a higher position, so we see Ek and Eg terms on the right side of this equation.
Substitute Energy Equations into Conservation Equations
½kx2 = ½mv’2 + mgh’
Reminders from Energy Notes: k is spring constant, and x is distance the spring is compressed.
Substitute Known Values
k = 1100 J/m2
x = 0.44 m
m = 0.80 kg
v’ = 13.1 m/s
h’ = ???
½(1100)(0.44 m)2 = ½(0.80)(13.1 )2 + (0.80)(10 )h’
106.48 J = 68.644 J + 8h’
37.836 J = 8h’
4.73 m = h’