91影视

The given data is listed below as-

• The distance covered by the package is, $\mathrm{s}=2.0\mathrm{m}$
• The mass of the package is, $\mathrm{m}=12\mathrm{kg}$
• The coefficient of kinetic friction between the package and the chute鈥檚 surface is ${\mathrm{渭}}_{\mathrm{k}}=0.40$

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

$$\begin{gathered} \mathbf{W}\mathbf{=}\stackrel{\mathbf{鈫�}}{\mathbf{F}}\mathbf{路}\stackrel{\mathbf{鈫�}}{\mathbf{s}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{Fs}\mathbf{}\mathbf{肠辞蝉蠒} \end{gathered}$$

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

The friction force is

$\mathrm{f}={\mathrm{渭}}_{\mathrm{k}}脳\mathrm{m}脳\mathrm{g}脳\mathrm{肠辞蝉胃}$

The work done by friction is

${\mathrm{W}}_{\mathrm{f}}={\mathrm{渭}}_{\mathrm{k}}脳\mathrm{m}脳\mathrm{g}脳\mathrm{肠辞蝉胃}脳鈭�\mathrm{s}$

Here, ${\mathrm{渭}}_{\mathrm{k}}$is the coefficient of friction between the package and the chute鈥檚 surface, m is the mass of the package, g is the gravitational constant, and $鈭�s$is the displacement.

For ${\mathrm{渭}}_{\mathrm{k}}=0.40,\mathrm{m}=12\mathrm{kg},鈭�\mathrm{s}=2.0\mathrm{m},\mathrm{胃}=53掳$and $\mathrm{g}=9.8\mathrm{m}.{\mathrm{s}}^{鈥�2}$

$$\begin{gathered} {\mathrm{W}}_{\mathrm{f}}=-{\mathrm{渭}}_{\mathrm{k}}脳\mathrm{m}脳\mathrm{g}脳\mathrm{肠辞蝉胃}脳鈭�\mathrm{s} \\ =-\left(0.4\right)脳\left(12\mathrm{kg}\right)脳\left(9.8\mathrm{m}.{\mathrm{s}}^{鈥�2}\right)脳\left(\cos 53掳\right)脳\left(2\mathrm{m}\right) \\ =-56.66\mathrm{J} \end{gathered}$$

Thus, the magnitude of work done on the package by friction is $-56.66\mathrm{J}$

The work done by gravity on the package is given by

$$\begin{gathered} {\mathrm{W}}_{\mathrm{g}}=\mathrm{m}脳\mathrm{g}脳\mathrm{蝉颈苍胃}脳鈭�\mathrm{s} \\ =\left(12\mathrm{kg}\right)脳\left(9.8\mathrm{m}.{\mathrm{s}}^{鈥�2}\right)脳\left(\sin 53掳\right)脳\left(2\mathrm{m}\right) \\ =187.97\mathrm{J} \end{gathered}$$

Thus, the work done by gravity on the package is given by $187.97\mathrm{J}$.

The work done is equal to zero because normal is perpendicular to the direction of motion.

Therefore.

${\mathrm{W}}_{\mathrm{n}}=0\mathrm{J}$

Thus, work done on the package by normal force is0 J.

Net work done on the package will be the sum of all the individual work

$$\begin{gathered} {\mathrm{W}}_{\mathrm{net}}={\mathrm{W}}_{\mathrm{f}}+{\mathrm{W}}_{\mathrm{g}}+{\mathrm{W}}_{\mathrm{n}} \\ =-56.66\mathrm{J}+187.97\mathrm{J}+0\mathrm{J} \\ =131.31\mathrm{J} \end{gathered}$$

Thus, the total work done on the package is 131.31 J.