91Ó°ÊÓ

The given data is listed below as-

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

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

$\begin{array}{rcl}\mathrm{W}& =& \stackrel{→}{\mathrm{F}}·\stackrel{→}{\mathrm{s}}\\ & =& \mathrm{Fs}\cos \mathrm{ϕ}\end{array}$

If F and s are in the same direction then localid="1668591080572" $\mathrm{ϕ}=0°$and if it is in opposite direction then $\mathrm{ϕ}=180°$.

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 $F_{rope}$and another is the component of gravitational force along the x-axis, which is as below

$F_1=w\cos \mathrm{θ}=mg \cos \mathrm{θ}$

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:

role="math" localid="1668591400876" $F_2=w\sin \mathrm{θ}=mg \sin \mathrm{θ}$

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

$w=Fs\cos \mathrm{Ï•}........\left(1\right)$

Therefore, $W_{rope}=F_{rope}s\cos \mathrm{Ï•}$

Here, $F_{rope}$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 $F_{rope}$and s.

For $F_{rope}=72N$, s = 0, and $\mathrm{ϕ}=0°$

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

Thus, the magnitude of work done on the carton by the ramp is

The work done by gravity on the carton is given by

$\begin{array}{rcl}{w}_g& =& F_{g}s\cos \mathrm{Ï•}\\ & =& mgs\cos \mathrm{Ï•}\end{array}$

The angle between $F_g$and s is$\mathrm{ϕ}=120°$

$\begin{array}{rcl}{w}_g& =& mgs\cos \mathrm{ϕ}\\ & =& 128N×5.2m×\cos 120°\\ & =& 128N×5.2m×\left(-0.5\right)\\ & =& -332.8N·m\end{array}$

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

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

$w_n=F_{n}s\cos \mathrm{Ï•}$

The angle between $F_n$and s is

$\begin{array}{rcl}{w}_n& =& F_{n}s\cos \mathrm{ϕ}\\ w_n& =& F_n×5.2×\cos 90°\\ & =& 0J\end{array}$

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

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

$\begin{array}{rcl}{W}_{net}& =& W_{rope}+W_g+W_n\\ & =& 374.4J-332.8J+0J\\ & =& 41.6J\end{array}$

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