Question 7.11: A 12.0 N force with a fixed orientation does work on a parti......
A 12.0 N force with a fixed orientation does work on a particle as the particle moves through the three-dimensional displacement \vec{d}=(2.00 \hat{i}-4.00 \hat{j}+3.00 \hat{k}) m. What is the angle between the force and the displacement if the change in the particle’s kinetic energy is (a) +30.0 J and (b) -30.0 J?
Using the work-kinetic energy theorem, we have
\Delta K=W=\vec{F}\cdot \vec{d}=F d \cos \phi.
In addition, F =12 N and d=\sqrt{(2.00 \,m )^2+(-4.00 \,m )^2+(3.00\, m )^2}=5.39 \,m.
(a) If ΔK = -30.0 J , then
\phi=\cos ^{-1}\left(\frac{\Delta K}{F d}\right)=\cos ^{-1}\left(\frac{30.0 \,J}{(12.0 \,N )(5.39 \,m )}\right)=62.3^{\circ}.
(b) ΔK = -30.0 J , then
\phi=\cos ^{-1}\left(\frac{\Delta K}{F d}\right)=\cos ^{-1}\left(\frac{-30.0 \,J}{(12.0 \,N )(5.39 \,m )}\right)=118^{\circ}.