# Is my textbook teaching an incorrect concept of Work?

I fear my textbook is teaching an incorrect concept of Work. I am very frustrated right now since I was struggling to understand the concept in the way that was explained in the textbook and my instincts tell me it is wrong. Also note that this is a online highschool course, so the textbooks are not published per se and it is common for these courses to be incorrect.

background information:

Here is the definition of work given in the textbook:

We say that work is done on an object when a force that is exerted on it causes a displacement.

I'll give two examples here as I consider both to be incorrect.

Textbook Example 1:

If a 10 N force causes an object to be displaced 5.0 m, what is the work done on the object?

W = Fd

W = 10 N x 5.0 m

W = 50 Nm

W = 50 J

Textbook Example 2:

A 0.0782 kg hockey puck was shot with an acceleration of 2.8 m/s2 over extremely fast ice (we will assume that it is frictionless). How much work was done when the puck passed the 12.0 m mark and the 112.0 m mark?

In order to determine the work done, we need to know the distance traveled (which we do) and the force (which we do not). However, because we know the mass of the puck and its acceleration, then we can determine the force and then, use this force to determine the work.

Fnet = ma

Fapplied = 0.0782 kg x 2.8 m/s2

F = 0.218 96 kg/m/s2

F = 0.218 96 N

F = 0.22 N

Now we know the force exerted on the puck, we can now determine the work done on the puck as it passes the 12.0 m mark.

W = Fd

W = 0.22 N x 12.0 m

W = 2.64 Nm

W = 2.6 J

We can also determine the work done on the puck as it passes the 112.0 m mark.

W = Fd

W = 0.22 N x 112.0 m

W = 24.64 Nm

W = 25 J

Notice that the distance the puck travels affects the amount of work done!

Why I think these are incorrect:

I believe these answers are incorrect because of what the unit Joules means to me. Newton meters would appear to be a meaningless unit of the distance in the calculation was the distance traveled by the object since friction is not considered and any distance might be possible that being the case. As well, it means that, as shown in the hockey puck problem, the resulting joules may be found to be thousands of Joules because the object is sliding on a frictionless surface, but there is no more a transfer of energy--and work IS defined as a transfer of energy. In the puck problem it doesn't make sense for the resulting Joules to be more after a longer distance since there is no more energy transferred. Once the puck is accelerated, it will remain at a constant velocity and its kinetic energy will be constant (since it is a frictionless surface). Therefore, the Joules of energy transfer should remain the same after the acceleration.

I think that Work should really be defined as: the distance the force was exerted over as opposed to the distance the object traveled. Please correct me if I'm wrong. If I'm right, I will send the link of this question to the head of the science department or school as it is a huge oversight to teach a completely wrong concept to students (including myself).

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You are right. Your definition is better, since it does not lead astray and their 2nd example is intended to demonstrate only one thing: they misunderstand the most basic concept of the physics! – Val Jul 13 '13 at 0:28