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.

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 $\mathrm{\varphi }=36.9°$

Workdone can be calculated using the forces exerted on the body and the total displacement of the body. Thereforeif a constant force$\overrightarrow{\mathbf{F}}$is appliedduring displacement$\overrightarrow{\mathbf{s}}$along a straight lineis 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 $\mathbf{\varphi }\mathbf{=}\mathbf{180}\mathbf{°}$.

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 isthe 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

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 $-324\mathrm{J}$.

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.