Question 7.14: Figure 7-27 shows an overhead view of three horizontal force......
Figure 7-27 shows an overhead view of three horizontal forces acting on a cargo canister that was initially stationary but now moves across a frictionless floor. The force magnitudes are F_1 = 3.00 N, F_2 = 4.00 N, and F_3 = 10.0 N, and the indicated angles are θ_2 = 50.0 and θ_3 = 35.0 .What is the net work done on the canister by the three forces during the first 4.00 m of displacement?
The forces are all constant, so the total work done by them is given by W = F_{net}Δx , where F_{net} is the magnitude of the net force and Δx is the magnitude of the displacement. We add the three vectors, finding the x and y components of the net force:
\begin{aligned}F_{\text{net}x}&=-F_1-F_2 \sin 50.0^{\circ}+F_3 \cos 35.0^{\circ}=-3.00 \,N -(4.00 \,N ) \sin 35.0^{\circ}+(10.0 \,N ) \cos 35.0^{\circ}\\&=2.13 \,N \\F_{\text{net}y}&=-F_2 \cos 50.0^{\circ}+F_3 \sin 35.0^{\circ}=-(4.00 \,N ) \cos 50.0^{\circ}+(10.0 \,N ) \sin 35.0^{\circ}\\&=3.17 \,N .\end{aligned}
The magnitude of the net force is
F_{\text{net}}=\sqrt{F_{\text{net}x}^2+F_{\text{net}y}^2}=\sqrt{(2.13 \,N )^2+(3.17\, N )^2}=3.82 \,N.
The work done by the net force is
W=F_{\text{net}}d=(3.82 \,N )(4.00\, m )=15.3 \,J
where we have used the fact that \vec{d}\| \vec{F}_{\text{net}} (which follows from the fact that the canister started from rest and moved horizontally under the action of horizontal forces — the resultant effect of which is expressed by \vec{F}_{\text{net}}).