91Ó°ÊÓ

The given data is listed below as-

• The distance covered by cartons along the surface of the ramp is, $\mathrm{s}=5.20\mathrm{m}$
• The mass of the baggage is, $\mathrm{w}=\mathrm{mg}=128\mathrm{N}$
• The coefficient of kinetic friction between the carton and the baggage ramp is${\mathrm{μ}}_{\mathrm{k}}=0$
• Force due to rope is$\mathrm{F}=72\mathrm{N}$
• The inclined angle between the baggage ramp and floor $\mathrm{Ï•}=30°$

Workdone can be calculated using the forces exerted on the body and the total displacement of the body. Thereforeif a constant force$\stackrel{\mathbf{→}}{\mathbf{F}}$is appliedduring displacement$\stackrel{\mathbf{→}}{\mathbf{s}}$along a straight lineis given by,

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

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

(a)

The free-body diagram for the carton is shown below:

Take the x-axis along the inclined baggage ramp y-axis perpendicular to the x-axis.

Along the x-axis, two forces are acting on the carton:

One is ${\mathrm{F}}_{\mathrm{rope}}$and another is the component of gravitational force along the x-axis, which is as below

${\mathrm{F}}_1=\mathrm{w}\mathrm{³¦´Ç²õθ}=\mathrm{mg³¦´Ç²õθ}$

The above two forces are opposite to each other.

Similarly along the y-axis also, two forces are acting on the carton.

One is normal force and another is a component of gravitational force along the y-axis which is as below:

${\mathrm{F}}_2=\mathrm{w}\mathrm{²õ¾\pm ²Ôθ}=\mathrm{mg²õ¾\pm ²Ôθ}$

Now, the work done on the carton by the rope will be

$\mathrm{w}=\mathrm{Fs}\mathrm{³¦´Ç²õÏ•}$ …â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(1)

Therefore, ${\mathrm{W}}_{\mathrm{rope}}={\mathrm{F}}_{\mathrm{rope}}\mathrm{s³¦´Ç²õÏ•}$

Here, role="math" localid="1664860920094" ${\mathrm{F}}_{\mathrm{rope}}$is the force due to rope, s is the distance covered by carton along the surface of the ramp, and $\mathrm{Ï•}$is the angle between ${\mathrm{F}}_{\mathrm{rope}}$and s

For ${\mathrm{F}}_{\mathrm{rope}}=72\mathrm{N},\mathrm{s}=5.20\mathrm{m}$,and $\mathrm{ϕ}=0°$

$$\begin{gathered} {\mathrm{W}}_{\mathrm{rope}}=72\mathrm{N}×5.2\mathrm{m}\cos 0° \\ =374.4\mathrm{N}·\mathrm{m} \\ =374.4\mathrm{J} \end{gathered}$$

Thus, the magnitude of work done on the carton by the ramp is $374.4\mathrm{J}$

(b)

The work done by gravity on the carton is given by

$$\begin{gathered} {\mathrm{w}}_{\mathrm{g}}={\mathrm{F}}_{\mathrm{g}}\mathrm{s³¦´Ç²õÏ•} \\ =\mathrm{mg}\mathrm{s}\mathrm{³¦´Ç²õÏ•} \end{gathered}$$

The angle between ${\mathrm{F}}_{\mathrm{g}}$ and s is $\mathrm{ϕ}=120°$

$$\begin{gathered} {\mathrm{w}}_{\mathrm{g}}=\mathrm{mg}\mathrm{s}\mathrm{³¦´Ç²õÏ•} \\ =128\mathrm{N}×5.2\mathrm{m}×\cos 120° \\ =128\mathrm{N}×5.2\mathrm{m}×\left(-0.5\right) \\ =-332.8\mathrm{N}·\mathrm{m} \end{gathered}$$

Thus, the work done by gravity on the carton is given by $-332.8\mathrm{J}$.

(c)

The work done by normal force on the carton is given by

${\mathrm{w}}_{\mathrm{n}}={\mathrm{F}}_{\mathrm{n}}\mathrm{s}\mathrm{³¦´Ç²õÏ•}$

The angle between ${\mathrm{F}}_{\mathrm{n}}$and s is $\mathrm{ϕ}=90°$

$$\begin{gathered} {\mathrm{w}}_{\mathrm{n}}={\mathrm{F}}_{\mathrm{n}}\mathrm{s}\mathrm{³¦´Ç²õÏ•} \\ {\mathrm{w}}_{\mathrm{n}}={\mathrm{F}}_{\mathrm{n}}×5.2×\cos 90° \\ =0\mathrm{J} \end{gathered}$$

Thus, work done on the carton by normal force is 0 J

(d)

Net work done on the carton will be the sum of all the individual work done and is given by

$$\begin{gathered} {\mathrm{W}}_{\mathrm{net}}={\mathrm{W}}_{\mathrm{rope}}+{\mathrm{W}}_{\mathrm{g}}+{\mathrm{W}}_{\mathrm{n}} \\ =374.4\mathrm{J}-332.8\mathrm{J}+0.\mathrm{J} \\ =41.6\mathrm{J} \end{gathered}$$

Thus, the total work done on the carton is 41.6 J.