Determining the Work to Accelerate a Package

This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See Figure 7.4.) As expected, the net work is the net force times distance.

The net force is the push force minus friction, or $F_{\text{net}}\text{= 120 N – 5}\text{.}\text{00 N = 115 N}$. Thus the net work is

This value is the net work done on the package. The person actually does more work than this, because friction opposes the motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force.

The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force and force of gravity are each perpendicular to the displacement, and therefore do no work.

The applied force does work.

The friction force and displacement are in opposite directions, so that $\theta =\text{180º}$, and the work done by friction is

So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively,

The total work done as the sum of the work done by each force is then seen to be

The calculated total work $W_{\text{total}}$ as the sum of the work by each force agrees, as expected, with the work $W_{\text{net}}$ done by the net force. The work done by a collection of forces acting on an object can be calculated by either approach.