91Ó°ÊÓ

The given data is listed below as-

• The displacement of the crate is, $\mathrm{s}=12\mathrm{m}$
• $\mathrm{Î\pm }=22°$east of North
• $\mathrm{β}=61°$South of West
• The force applied by Paul, $\mathrm{F}=48\mathrm{N}$

Consider a particle with a constant force $\stackrel{\mathbf{→}}{\mathbf{F}}$during a linear displacement $\stackrel{\mathbf{→}}{\mathbf{d}}$. The work done on that particle is given by-

$$\begin{gathered} \mathbf{W}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{·}\stackrel{\mathbf{→}}{\mathbf{d}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{Fd}\mathbf{}\mathbf{³¦´Ç²õÏ•} \end{gathered}$$

If F and d are in the same direction then $\mathbf{ϕ}\mathbf{=}\mathbf{0}$ and if it is in opposite direction then $\mathbf{ϕ}\mathbf{=}\mathbf{180}\mathbf{°}$.

The free-body diagram for the crate is as shown below:

The work done on the crate by Paul is given by

Here, F is the force applied by Paul, d is the displacement of the crate, and $\mathrm{θ}$is the angle between force and direction of motion.

For $\mathrm{m}=48\mathrm{N},\mathrm{d}=12\mathrm{m},$and $\mathrm{θ}=\mathrm{Î\pm }+90°+\mathrm{β}=173°$

$$\begin{gathered} \mathrm{W}=\mathrm{Fd}\cos \mathrm{θ} \\ =48\mathrm{N}×12\mathrm{m}×\cos 173° \\ =-571.7\mathrm{J} \end{gathered}$$

Thus, the magnitude of work done on the crate by Paul’s force during a displacement of the carte is $-571.7\mathrm{J}$