Q13E - University Physics with Modern Physics

Question

 A large crate sits on the floor of a warehouse. Paul and Bob also apply constant horizontal forces to the crate. The force applied by Paul has magnitude 48.0 N and direction $\mathbf{61}\mathbf{.}\mathbf{0}\mathbf{°}$ south of west. How much work does Paul’s force do during a displacement of the crate that is 12.0 m in the direction  $\mathbf{22}\mathbf{°}$ east of north?

Step-by-Step Solution

Answer

 $-571.7 \mathrm{J}$

Step 1: Identification of the given data

The given data is listed below as-

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

Step 2: Significance of the work done

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

$$\begin{gathered} \mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{·}\overrightarrow{\mathbf{d}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{Fd}\mathbf{ }\mathbf{cos\varphi } \end{gathered}$$

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

Step 3: Determination of work done on the crate by Paul

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

  

The work done on the crate by Paul is given by 

$$\begin{gathered} \mathrm{W}=\overrightarrow{\mathrm{F}}·\overrightarrow{\mathrm{d}} \\ =\mathrm{Fd} \cos \mathrm{\theta } \end{gathered}$$ 

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

For $\mathrm{m}=48 \mathrm{N}, \mathrm{d}=12 \mathrm{m},$ and  $\mathrm{\theta }=\mathrm{\alpha }+90°+\mathrm{\beta }=173°$

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

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