• If force and displacement are both in the same direction, the work done is the product of the magnitudes of force and displacement.
• If force and displacement are at an angle to one another, you need to calculate the component of the force that points in the direction of the displacement, or the component of the displacement that points in the direction of the force. The work done is the product of the one vector and the component of the other vector.
• If force and displacement are perpendicular, no work is done.

    Because of the way work is defined in physics, there are a number of cases that go against our everyday intuition. Work is not done whenever a force is exerted, and there are certain cases in which we might think that a great deal of work is being done, but in fact no work is done at all. Let’s look at some examples that might be tested on SAT II Physics:

• You do work on a 10 kg mass when you lift it off the ground, but you do no work to hold the same mass stationary in the air. As you strain to hold the mass in the air, you are actually making sure that it is not displaced. Consequently, the work you do to hold it is zero.
• Displacement is a vector quantity that is not the same thing as distance traveled. For instance, if a weightlifter raises a dumbbell 1 m, then lowers it to its original position, the weightlifter has not done any work on the dumbell.
• When a force is perpendicular to the direction of an object’s motion, this force does no work on the object. For example, say you swing a tethered ball in a circle overhead, as in the diagram below. The tension force, T, is always perpendicular to the velocity, v, of the ball, and so the rope does no work on the ball.

    Since the gravitational force of –mg is in the same direction as the water balloon’s displacement, –h, the work done by the gravitational force on the ball is the force times the displacement, or W = mgh, where g = –9.8 m/s².

 

    How much work is done by gravity if the balloon is thrown horizontally from a height h with an initial velocity of v0?

 

    The gravitational force exerted on the balloon is still –mg, but the displacement is different. The balloon has a displacement of –h in the y direction and d (see the figure below) in the x direction. But, as we recall, the work done on the balloon by gravity is not simply the product of the magnitudes of the force and the displacement. We have to multiply the force by the component of the displacement that is parallel to the force. The force is directed downward, and the component of the displacement that is directed downward is –h. As a result, we find that the work done by gravity is mgh, just as before.