Q12E - University Physics with Modern Physics

Question

A boxed 10.0-kg computer monitor is dragged by friction 5.50 m upward along a conveyor belt inclined at an angle of  $\mathbf{36}\mathbf{.}\mathbf{9}\mathbf{°}$ above the horizontal. If the monitor’s speed is a constant 2.10 cm/s, how much work is done on the monitor (a) by friction, (b) by gravity, and (c) the normal force of the conveyor belt.

Step-by-Step Solution

Answer

(a)  ${\mathrm{W}}_{\mathrm{f}}=324 \mathrm{J}$

 

(b)  ${\mathrm{W}}_{\mathrm{g}}=-324 \mathrm{J}$

 

(c)  ${\mathrm{W}}_{\mathrm{n}}=0 \mathrm{J}$

Step 1: Identification of the given data

The given data is listed below as-

• The distance covered by the monitor when it is dragged upward is, $\mathrm{s}=5.50 \mathrm{m}$  
• The mass of the monitor is, $\mathrm{m}=10 \mathrm{kg}$  
• The angle of the conveyor belt with the horizontal $\mathrm{\varphi }=36.9°$  

Step 2: Significance of the work done

Work done can be calculated using the forces exerted on the body and the total displacement of the body. Therefore if a constant force $\overrightarrow{\mathbf{F}}$  is applied during displacement  $\overrightarrow{\mathbf{s}}$ along a straight line is given by,

$$\begin{gathered} \mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{·}\overrightarrow{\mathbf{s}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{Fs}\mathbf{ }\mathbf{cos\varphi } \end{gathered}$$ 

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

Step 3: Determination of work done on the monitor by friction (a)

It is given that the monitor’s speed is constant, therefore it can be concluded that the total force acting on the monitor is 0 according to Newton’s law. 

Therefore, the sum of all the forces in the direction of motion and perpendicular to it is zero.

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

The work done on the monitor due to friction is given by

${\mathrm{W}}_{\mathrm{f}}={\mathrm{F}}_{\mathrm{f}} \mathrm{s}$  

Where, ${\mathrm{F}}_{\mathrm{f}} =\mathrm{mg} \mathrm{sin\theta }$  

Here, m is the mass of the monitor, and g is the gravitational constant.

 For $\mathrm{m}=10 \mathrm{kg}, \mathrm{g}=9.8 \mathrm{m}.{\mathrm{s}}^{-2},$ and  $\mathrm{\theta }=36.9°$

$$\begin{gathered} {\mathrm{F}}_{\mathrm{f}}=\mathrm{mg} \sin \mathrm{\theta } \\ =10\times 9.8\times \sin \left(36.9°\right) \\ =58.84 \mathrm{N} \end{gathered}$$

 

Now,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{f}}={\mathrm{F}}_{\mathrm{f}} \mathrm{s} \\ =58.84 \mathrm{N}\times 5.5 \mathrm{m} \\ =324 \mathrm{J} \end{gathered}$$

  

Thus, the magnitude of work done on the monitor by friction is 342 J 

Step 4: Determination of work done on the monitor due to gravity (b)

The component of gravity parallel to the direction of motion is given by  $\mathrm{mg} \sin \mathrm{\theta }$ and is in the opposite direction to it.

The work done by gravity on the monitor is given by

$$\begin{gathered} {\mathrm{W}}_{\mathrm{g}}=-\mathrm{mg} \mathrm{s} \sin \mathrm{\theta } \\ =-{\mathrm{F}}_{\mathrm{f}} \mathrm{s} \\ =-324 \mathrm{J} \end{gathered}$$

 

Thus, the work done by gravity on the monitor is given by $-324 \mathrm{J}$ .

Step 5: Determination of work done on the monitor by normal force of the conveyor belt (c)

The conveyor belt does not do any work because the normal force of the conveyor belt is perpendicular to the direction of motion. Therefore, ${\mathrm{w}}_{\mathrm{n}}=0$ 

Thus, work done by the normal force of the conveyor belt on the monitor is 0 J.