How to Calculate Total Work

“Work” has a specific meaning in physics, as discussed in Halliday and Resnick’s “Fundamentals of Physics.” It’s the product of the distance a body moves and the force in the direction of motion that caused that movement. Or more imprecisely, work is force times distance. So if you’re holding a book stationary in the air, you’re not doing work on the book, though you yourself are clearly burning energy to maintain muscle tension. If you’re pulling a sled, the force you exert may have some upward component, but that doesn’t count toward the work you do on the sled. To calculate the component of the force tangent to the direction of motion, you’d perform what’s called the “dot product,” which is a vector operation that is usually denoted with a middle-level dot, but which will be denoted here with a period.

Instructions

1. Draw a diagram in which the force and vector of displacement of a body is articulated. <br /><br />For example, suppose you are pulling a sled with a force F= 10 newtons (about 2.2 pounds) that is to the left and upward, at a 30-degree angle from horizontal. Then the sled’s displacement vector r points to the left. Suppose also that the sled is to experience this force F through a horizontal displacement of 10 meters and will experience no vertical displacement, since the upward component won’t be enough to lift the sled off the ground any distance. So the vector r has a direction to the left and a magnitude of 10 meters.

2. Take the dot product of vectors F and r, using the property that for vectors a and b, their dot product is the product of their magnitudes times the cosine of the angle between them.<br /><br />So continuing with the example above, the dot product is F.r = magnitude(F) x magnitude(r) x cos ? = 10N x 10m x cos 30 = 100Nm x ?3/2 = 86.60 Joules. (A newton-meter equals a joule, the SI unit for energy.)

3. Equate the result in Step 2 with the work expended on the body, rounding to the appropriate number of significant figures.<br /><br />So 86.60 joules is the work expended on the sled. Even though additional energy was expended to apply vertical force on the sled as well, the sled didn’t move vertically, so that energy doesn’t translate into actual work on the sled itself.

Tips & Warnings

• To understand more intuitively how pushing on a body doesn’t constitute work on it, consider a pedestal that supports a vase, bust or roof. The pedestal works to oppose gravity, but the bust experiences no work on it because the gravitational force and pedestal’s force cancel each other out.<br /><br />Account for variation in vectors F or r by integrating the dot product as a line integral, ?F.dr, using vector calculus. Here, dr is a differential vector. How to replace the dot with a cosine function and replace the dr with dx or dy in order to carry out the integration would depend on the physical situation being measured.

References

• Argonne National Lab: Work Define and Muscle Energy
• Math Help: The Scalar or Dot Product

Resources

• Physics Forums: Work Done Using Dot Product