To make this process easy to understand, let’s follow along with an example problem. Say that a toy train car is being pulled directly forward by the train in front of it. In this case, both the force vector and the direction of the train’s motion point the same way — forward. In the next few steps, we’ll use this information to help find the work done on the object.
Note that measures of distance must be in meters for the work formula. In our toy train example, let’s say that we’re finding the work performed on the train as it travels along the track. If it starts at a certain point and ends at a spot about 2 meters (6. 6 ft) up the track, we can use 2 meters (6. 6 ft) for our “D” value in the formula.
Note that measures of force must be in newtons for the work formula. In our example, let’s say that we don’t know the magnitude of the force. However, let’s say that we do know that the toy train has a mass of 0. 5 kilograms and that the force is causing it to accelerate at a rate of 0. 7 meters/second2. In this case, we can find the magnitude by multiplying M × A = 0. 5 × 0. 7 = 0. 35 Newtons.
It’s time to solve our example problem. With a value for force of 0. 35 Newtons and a value for displacement of 2 meters (6. 6 ft), our answer is a single multiplication problem away: 0. 35 × 2 = 0. 7 joules. You may have noticed that, in the formula provided in the intro, there’s an additional piece to the formula: Cosine(θ). As discussed above, in this example, the force and the direction of motion are in the same direction. This means the angle between them is 0o. Since Cosine(0) = 1, we don’t need to include it — we’re just multiplying by 1.
Note that joules also has an alternate definition — one watt of power radiated over one second. [7] X Research source See below for a more detailed discussion of power and its relationship to work.
Let’s look at another example problem. In this case, let’s say that we’re pulling a toy train forward as in the example problem above, but that this time we’re actually pulling upward at a diagonal angle. In the next step, we’ll take this into account, but for now, we’ll stick to the basics: the train’s displacement and the magnitude of the force acting on it. For our purposes, let’s say that the force has a magnitude of 10 newtons and that it’s moved the same 2 meters (6. 6 ft) forward as before.
In our example problem, let’s say that the force is being applied about 60o above the horizontal. If the train is still moving directly forward (that is, horizontally), the angle between the force vector and the train’s motion is 60o.
Let’s solve our example problem. Using a calculator, we find that the cosine of 60o is 1/2. Plugging this into the formula, we can solve as follows: 10 newtons × 2 meters (6. 6 ft) × 1/2 = 10 joules.
For example, let’s say that we know that our train is being pulled with 20 newtons of force at a diagonal angle over 5 meters (16. 4 ft) of track to perform 86. 6 joules of work. However, we don’t know the angle of the force vector. To solve for the angle, we’ll just isolate that variable and solve as follows: 86. 6 = 20 × 5 × Cosine(θ) 86. 6/100 = Cosine(θ) Arccos(0. 866) = θ = 30o
For instance, for the example problem in the step above, let’s say that it took 12 seconds for the train to move 5 meters (16. 4 ft). In this case, all we need to do is divide the work done to move it 5 meters (86. 6 joules) by 12 seconds to find our answer for power: 86. 6/12 = ‘7. 22 watts.
For instance, for the example problem in the two steps above, let’s say that the train initially had a total mechanical energy of 100 joules. Since the force in the problem is pulling the train in the direction that it’s already traveling, it’s positive. In this case, the train’s final energy is TMEi + Wnc = 100 + 86. 6 = 186. 6 joules. Note that non-conservative forces are forces whose power to affect an object’s acceleration depends on the path taken by the object. Friction is a good example — an object pushed over a short, direct path will feel the effects of friction for a short while, while an object pushed over a long, meandering path to the same ending location will feel more friction overall.
