Q10E - University Physics with Modern Physics

Question

A 12.0-kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0°

  below the horizontal. The coefficient of kinetic friction between the package and the chute’s surface is 0.40. Calculate the work done on the package by (a) friction, (b) gravity, and (c) the normal force, (d) What is the net work done on the package.

Step-by-Step Solution

Answer

(a)  ${\mathrm{W}}_{\mathrm{f}}=-56.66 \mathrm{J}$

 

(b)  ${\mathrm{W}}_{\mathrm{g}}=187.97 \mathrm{J}$

 

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

 

(d)  ${\mathrm{W}}_{\mathrm{net}}=131.31 \mathrm{J}$

Step 1: Identification of the given data

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’s surface is ${\mathrm{\mu }}_{\mathrm{k}}=0.40$  

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 package by friction (a)

The friction force is 

$\mathrm{f}={\mathrm{\mu }}_{\mathrm{k}}\times \mathrm{m}\times \mathrm{g}\times \mathrm{cos\theta }$  

The work done by friction is 

${\mathrm{W}}_{\mathrm{f}}={\mathrm{\mu }}_{\mathrm{k}}\times \mathrm{m}\times \mathrm{g}\times \mathrm{cos\theta }\times ∆\mathrm{s}$ 

Here, ${\mathrm{\mu }}_{\mathrm{k}}$ is the coefficient of friction between the package and the chute’s surface, m is the mass of the package, g is the gravitational constant, and $∆s$ is the displacement.

 For ${\mathrm{\mu }}_{\mathrm{k}}=0.40, \mathrm{m}=12\mathrm{kg}, ∆\mathrm{s}=2.0 \mathrm{m}, \mathrm{\theta }=53°$ and $\mathrm{g}=9.8\mathrm{m}.{\mathrm{s}}^{‐2}$ 

$$\begin{gathered} {\mathrm{W}}_{\mathrm{f}}=-{\mathrm{\mu }}_{\mathrm{k}}\times \mathrm{m}\times \mathrm{g}\times \mathrm{cos\theta }\times ∆\mathrm{s} \\ =-\left(0.4\right)\times \left(12 \mathrm{kg}\right)\times \left(9.8\mathrm{m}.{\mathrm{s}}^{‐2}\right)\times \left(\cos 53°\right)\times \left(2\mathrm{m}\right) \\ =-56.66 \mathrm{J} \end{gathered}$$

 

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

Step 4: Determination of work done on the package by gravity (b)

The work done by gravity on the package is given by

$$\begin{gathered} {\mathrm{W}}_{\mathrm{g}}=\mathrm{m}\times \mathrm{g}\times \mathrm{sin\theta }\times ∆\mathrm{s} \\ =\left(12 \mathrm{kg}\right)\times \left(9.8 \mathrm{m}.{\mathrm{s}}^{‐2}\right)\times \left(\sin 53°\right)\times \left(2 \mathrm{m}\right) \\ =187.97 \mathrm{J} \end{gathered}$$

 

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

Step 5: Determination of work done on the package by the normal force (c)

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 is 0 J.

Step 6: Determination of net work done on the package (d)

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.