During a storm, a crate of crepe is sliding across a slick, oily parking lot through a displacement `vec(d)=(-3.0m)hat(i)` while a steady wind pushes against the crate with a force `vec(F)=(2.0N)hat(i)+(-6.0N)hat(j)`. The situation and coordinate axes are shown in Fig. 8-6.
(a) How much work does this force do on the crate during the displacement ?

Figure 8-6 Force `vec(F)` slows a crate during displacement `vec(d)`.

Explanation

KEY IDEAS
Because we can treat the crate as a particle and because the wind force is constant ("steady") in both magnitude and direction during the displacement, we can use either Eq. 8-7 `(W= Fd cos phi)` or Eq. 8-8 `(W=vec(F)*vec(d))` to calculate the work. Since we know `vec(F)` and `vec(d)` in unit-vector notation, we choose Eq. 8-8.
Calculations: We write
`W=vec(F)*vec(d)=[(2.0N)hat(i)+(-6.0N)hat(j)]*[(-3.0 m)hat(i)]`.
Of the possible unit-vector dot products, only `hat(i)*hat(i), hat(j)*hat(j)`, and `hat(k)*hat(k)` are nonzero. Here we obtain
`W=(2.0N)(-3.0m)hat(i)*hat(i)+(-6.0N)(-3.0m)hat(j)*hat(i)`
`=(-6.0J)(1)+0= - 6.0J`.
Thus, the force does a negative 6.0 J of work on the crate, transferring 6.0 J of energy from the kinetic energy of the crate.