While taking intro science classes I've notice a lot of people who don't have any system for solving basic science formulas. Once one develops a system for solving these equations they become very easy. Indeed, it is a criticism of many science classes that simply solving these equations blindly doesn't teach one anything about science. I'll leave that debate for somewhere else, and instead go over the process of solving these types of problems.
I'd also like to note that solving simple physics formulas isn't only useful for those in science classes. Over on the Physics Forums I see many people asking questions that they could solve themselves by simply Googling the formulas and then plug and chug away.
I'd also like to note that solving simple physics formulas isn't only useful for those in science classes. Over on the Physics Forums I see many people asking questions that they could solve themselves by simply Googling the formulas and then plug and chug away.
Before we begin, you should be able to manipulate algebraic formulas to solve for different variables. Given:
$$$a+b=\frac{c}{d}$$$
you should be able to solve for a, b, c, or d. For example, $$$d = \frac{c}{a+b}$$$. If you're unsure about how to do this Khan Academy is a great site covering tons of math and science at all levels.
Second, it would behoove you to familiarize yourself with the common variables used in whatever area you are working. Usually, these are pretty obvious (e.g. m for mass, v for velocity), but sometimes the obvious choice is already taken and something else must be used.
For the first example here are some formulas involving energy:
E = mhg
E = 1/2 m v2
Where:
E is energy, m is mass, h is height, g is gravitation acceleration = 9.8 m/s, and v is velocity.
Q: What is the velocity of an object with a mass of 500 kg and a kinetic energy of 100,000 J?
Begin by identifying the variables given, as well as the one asked for. We identify m = 500 kg, E = 100,000 J, and v = ? (we are looking for it). Now we must find a formula that has these variables. The second formula (E = 1/2 m v2) has all these variables. However, we need to solve for v. We can either solve for v before (while it is only variables) or after (while is is all numbers) we substitute numbers for variables. Which is easier will depend on the problem and person. Solving first has the advantage of avoiding some messy arithmetic, particularly on tests where the numbers are likely to cancel out nicely. It also only has to be done once for a given variable. So, if you have several problems all asking for the same variable it will make sense to solve for that variable once.
Getting back to this problem, I would solve for v first:
$$E = \frac{1}{2} m v^2$$
$$\frac{2 \cdot E}{m} = v^2$$
$$\sqrt{\frac{2 \cdot E}{m}}=v$$ (Ignoring the negative root)
Now we replace the variables with the known values:
$$\sqrt{\frac{2 \cdot 100,000\,\mathrm{J}}{500\,\mathrm{kg}}}=v$$
And now we solve the arithmetic:
$$v = 20\,\mathrm{\frac{m}{s}}$$
I'd like to mention something about units. You could simply recognize that meters per second is the expected units for velocity and assume that is what your result is. However, you can also solve units in the same way as you solve numbers. In this case we need to know that J = $$$\mathrm{\frac{kg \cdot m^2}{s^2}}$$$. Dividing J by kg is the same thing as multiplying it by 1/kg. Hence, the above formula with numbers removed gives:
$$\sqrt{\mathrm{\frac{kg \cdot m^2}{kg \cdot s^2}}}=\mathrm{v}$$
kg/kg cancels out giving:
$$\sqrt{\mathrm{\frac{m^2}{s^2}}}=\mathrm{v}$$
And taking the square root gives the expected m/s.
For the first example here are some formulas involving energy:
E = mhg
E = 1/2 m v2
Where:
E is energy, m is mass, h is height, g is gravitation acceleration = 9.8 m/s, and v is velocity.
Q: What is the velocity of an object with a mass of 500 kg and a kinetic energy of 100,000 J?
Begin by identifying the variables given, as well as the one asked for. We identify m = 500 kg, E = 100,000 J, and v = ? (we are looking for it). Now we must find a formula that has these variables. The second formula (E = 1/2 m v2) has all these variables. However, we need to solve for v. We can either solve for v before (while it is only variables) or after (while is is all numbers) we substitute numbers for variables. Which is easier will depend on the problem and person. Solving first has the advantage of avoiding some messy arithmetic, particularly on tests where the numbers are likely to cancel out nicely. It also only has to be done once for a given variable. So, if you have several problems all asking for the same variable it will make sense to solve for that variable once.
Getting back to this problem, I would solve for v first:
$$E = \frac{1}{2} m v^2$$
$$\frac{2 \cdot E}{m} = v^2$$
$$\sqrt{\frac{2 \cdot E}{m}}=v$$ (Ignoring the negative root)
Now we replace the variables with the known values:
$$\sqrt{\frac{2 \cdot 100,000\,\mathrm{J}}{500\,\mathrm{kg}}}=v$$
And now we solve the arithmetic:
$$v = 20\,\mathrm{\frac{m}{s}}$$
I'd like to mention something about units. You could simply recognize that meters per second is the expected units for velocity and assume that is what your result is. However, you can also solve units in the same way as you solve numbers. In this case we need to know that J = $$$\mathrm{\frac{kg \cdot m^2}{s^2}}$$$. Dividing J by kg is the same thing as multiplying it by 1/kg. Hence, the above formula with numbers removed gives:
$$\sqrt{\mathrm{\frac{kg \cdot m^2}{kg \cdot s^2}}}=\mathrm{v}$$
kg/kg cancels out giving:
$$\sqrt{\mathrm{\frac{m^2}{s^2}}}=\mathrm{v}$$
And taking the square root gives the expected m/s.
For a second example I'll be using an electricity question. Here are several formulas in this area:
$$F=\frac{k \cdot q_1 \cdot q_2}{r^2}$$
$$V=IR$$
$$R=\frac{\rho \cdot L}{A}$$
$$E=\frac{k \cdot q}{r^2}$$
$$V=\frac{w}{q}$$
$$A=π \cdot r^2$$
Where:
F is force from the charge.
k is Coulomb's constant = 9 x 109 $$$\mathrm{\frac{N \cdot m^2}{C^2}}$$$
qn is the charge of particle n.
r is the radius, or distance between the two charges.
V is voltage aka electrical potential difference
I is electrical current, measured in amps
R is resistance
ρ (greek letter rho) is resistivity of a material
L is length
A is cross sectional area
E is electrical field strength (volts/meter)
w is work
Here is a list of ρ (rho) values for common materials:
Q: A copper wire is 40 cm long and 1 cm in diameter. What is the resistance of the wire?
Where:
F is force from the charge.
k is Coulomb's constant = 9 x 109 $$$\mathrm{\frac{N \cdot m^2}{C^2}}$$$
qn is the charge of particle n.
r is the radius, or distance between the two charges.
V is voltage aka electrical potential difference
I is electrical current, measured in amps
R is resistance
ρ (greek letter rho) is resistivity of a material
L is length
A is cross sectional area
E is electrical field strength (volts/meter)
w is work
Here is a list of ρ (rho) values for common materials:
| Material | ρ [Ω·m] at 20 °C |
| Aluminium | 2.82×10-8 |
| Calcium | 3.36×10-8 |
| Copper | 1.68×10-8 |
| Gold | 2.44×10-8 |
| Iron | 1.0×10-7 |
| Lithium | 9.28×10-8 |
| Nickel | 6.99×10-8 |
| Silver | 1.59×10-8 |
| Tungsten | 5.60×10-8 |
| Zinc | 5.90×10-8 |
Q: A copper wire is 40 cm long and 1 cm in diameter. What is the resistance of the wire?
Begin by identifying all the variables given, as well as the one being asked for. Since all the formulas use standard units (meters, kilograms) I find it best to just immediately convert these into the standard units.
L = .4 m
d = .01 m
ρ = 1.68×10-8 Ω·m (value for copper taken from above chart)
R = ? (this is what the question is looking for)
Now we look at our available formulas and find the one which uses those variables. In this case, none only have those variables. However, $$$R=\frac{\rho \cdot L}{A}$$$ is only missing one, A (cross sectional area). Thus, we must solve a sub problem first. We see there is a formula for cross sectional area provided: $$$A=π \cdot r^2$$$, this only requires radius, which we know is half of diameter (if you wish, you can pretend d = 2r was provided and do a second sub problem to find r). Using the area formula we find that:
A = 7.85 x 10-5 m2
With the above list of variables, we can now solve for the original unknown (R).
First rewrite the original formula:
$$R=\frac{\rho \cdot L}{A}$$
Then substitute in the known variables:
$$R=\frac{1.68×10^{-8} \Omega\,\mathrm{m} \cdot 0.4\,\mathrm{m}}{7.85 × 10^{-5}\,\mathrm{m}^{2}}$$
Then solve the numbers and units separately:
$$R= 8.56×10^{-5} \frac{\Omega\,\mathrm{m}^2}{\mathrm{m}^2}$$
Units cancel the same way numbers do. Thus, m2/m2 cancels out and we are left with just Ω as our units, which is what we expect. It is easier to just drop all the units and then assume the final units are what they are supposed to be. However, actually solving the units provides an important check. If you are unsure if you are correctly remembering a formula simply solving with only units can tell you if its wrong. In addition, solving the units while doing the math will help show simple errors.
I could continue giving examples, however I think the basic process is pretty easy to follow (also laziness). Just identify the variables. Find a formula with the correct variables in it. Plug in the numbers and solve. I suggest you simply think of things to calculate then Google to find the formulas.
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