13 Doing Work
Lawrence Davis
We started the previous unit with a discussion of Jolene’s motion during a shift on the medical floor of a hospital, including all the starts and stops that she makes. When Jolene is standing still she has zero kinetic energy. As she takes a step to begin walking she now has kinetic energy. Jolene had to supply that energy from within herself. When Jolene comes to a stop her kinetic energy is transferred to thermal energy by friction. When she begins walking again she will need to supply the new kinetic energy all over again. Even if Jolene walks continuously, every step she takes involves two inelastic collisions (the push-off and the landing) so kinetic energy is constantly being transferred to thermal energy. To stay in motion Jolene has to re-supply that kinetic energy. Walking around all shift uses up Jolene’s stored energy and that is why she gets tired.
Work
The amount of energy transferred from one form to another and/or one object to another is called the work. Doing work is the act of transferring that energy. Doing work requires applying a force over some distance. The sign of the work done on an object determines if energy is transferred in or out of the object. For example, the athlete on the right is doing positive work on the pole because he is applying a force in the same direction as the pole’s motion. That will tend to speed up the pole and increase the kinetic energy of the pole. The athlete on the left is doing negative work on the pole because the force he applies tends to decrease the energy of the pole.
The positive or negative sign of the work refers to energy transferring in or out of an object rather than to opposite directions in space so work is not a vector and we will not make it bold in equations.
Calculating Work
The actual amount of work done is calculated from a combination of the average force and the distance over which it is applied, and the angle between the two:
[latex]\begin{equation} W = Fdcos\theta \end{equation}[/latex]
Everyday Example: Lifting a Patient
Jolene works with two other nurses to lift a patient that weighs 867 N (190 lbs) a distance of 0.5 m straight up. How much work did she do? Assuming Jolene lifted 1/3 of the patient weight, she had to supply an upward force of 289 N. The patient also moved upward, so the angle between force and motion was 0°. Entering these values in the work equation:
[latex]\begin{equation*} W = Fdcos\theta = (289{N})(0.5{m})cos(0^{\circ}) = 144{Nm} \end{equation*}[/latex]
We see that work has units of Nm, which are called a Joules (J). Work and all other forms of energy have the same units because work is an amount of energy, but work is not a type of energy. When calculating work the costheta accounts for the force direction so we only use the size of the force (F) in the equation, which is why we have not made force bold in the work equation.
The [latex]cos\theta[/latex] in the work equation automatically tells us whether the work is tr