Cracking the SAT Physics Subject Test

(a) How much work is done by the normal force?

(b) How much work is done by the friction force?

Here’s How to Crack It

(a) Clearly, the normal force is not parallel to the motion, so we use the

general definition of work. Since the angle between FN and d is 90° (by the

definition of normal) and cos 90° = 0, the normal force does zero work.

(b) The frictional force is antiparallel to the intended direction of motion.

When we lift the object in the position shown, we also give the object a slight

vertical lift, which decreases the normal force. The frictional force is equal to

the coefficient of friction times the normal force, not the weight.

In this case,

Weight = Fy, vertical component + Normal Force

N = mg − FT sin θ

Ff = (mg − FT sinθ)μ

Because friction opposes the direction of motion, the work is

negative (cos 180 = −1), so:

W = −Ff d

W = (mg − FT sinθ)μd

The two previous examples show that work—which, as we said, is a scalar
quantity—may be positive, negative, or zero. If the angle between F and d (θ) is
less than 90°, then the work is positive (because cosθ is positive in this case); if θ
= 90°, the work is zero (because cos 90° = 0); and if θ > 90°, then the work is
negative (because cos θ is negative). In other words, if a force helps the motion, the
work done by the force is positive, but if the force opposes the motion, then the
work done by the force is negative.

For situations where θ is something other than 0°, 90°, or 180°, it is sometimes
useful to break the force into components, F⊥ and F||.