Q1E - University Physics with Modern Physics

Question

You push your physics book 1.5 m along a horizontal table-top with a horizontal push of 2.40 N while the opposing force of friction is 0.600 N. How much work does each of the following forces do on the book: (a) your 2.40 N- push, (b) the friction force, (c) txhe normal force from the tabletop, and (d) gravity (e) What is the net work done on the book.

Step-by-Step Solution

Answer

$$\begin{gathered} \left(\mathrm{a}\right) {\mathrm{W}}_{\mathrm{push}}=3.6 \mathrm{J} \\ \left(\mathrm{b}\right) {\mathrm{W}}_{\mathrm{f}}=-0.9 \mathrm{J} \\ \left(\mathrm{c}\right) {\mathrm{W}}_{\mathrm{N}}=0 \mathrm{J} \\ \left(\mathrm{d}\right) {\mathrm{W}}_{\mathrm{g}}=0 \mathrm{J} \\ \left(\mathrm{e}\right) {\mathrm{W}}_{\mathrm{net}}=2.7 \mathrm{J} \end{gathered}$$

Step 1: Identification of the given data

The given data is listed below as-

• The distance covered by a book along a horizontal tabletop  is, s = 1.5 m   
• Applied horizontal push is,  F = 2.4 N 
• The frictional force, ${\mathrm{F}}_{\mathrm{f}}=0.6 \mathrm{N}$ 

Step 2: Significance of the work done

Work done on a particle by a constant force   during a linear displacement   is given by 

$$\begin{gathered} \mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{s}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{Fcos\varphi } \end{gathered}$$ …………..(1)

If F and s are in the same direction then $\mathrm{\varphi }=0$ and if it is in opposite direction then

$\mathrm{\varphi }=180°$ .

Step 3: Determination of work done by various forces on the book

(a) 

Work done on the book by F = 2.4 N push can be expressed by equation (1) such that,

$\mathrm{W}=\mathrm{Fscos\varphi }$ 

Since F and s both are in the same direction, $\mathrm{\varphi }=0$

Therefore, W = F s cos0

Here, F is applied horizontal push and s is the distance covered by book.

$$\begin{gathered} \mathrm{For} \mathrm{F}=2.4 \mathrm{N}, \mathrm{s}=1.5 \mathrm{m} \\ {\mathrm{W}}_{\mathrm{push}}=\mathrm{Fscos}0 \\ =2.4\mathrm{N}\times 1.5\mathrm{m}\times 1 \\ =3.6\mathrm{J} \end{gathered}$$

(b)

Work done on the book by the frictional force ${\mathrm{F}}_{\mathrm{f}}=0.6\mathrm{N}$ will be

${\mathrm{W}}_{\mathrm{f}}=\mathrm{Fs} \mathrm{cos\varphi }$ 

Since F and s both are in opposite directions, $\mathrm{\varphi }=180°$

Therefore,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{f}}=\mathrm{Fs} \cos 180° \\ =-\mathrm{Fs} \end{gathered}$$

Here, F is the frictional force and s is the distance covered by the book

$$\begin{gathered} \mathrm{For} \mathrm{F}=0.6\mathrm{N}, \mathrm{s}=1.5\mathrm{m} \\ \mathrm{W}=\mathrm{Fscos}180° \\ =0.6\mathrm{N}\times 1.5\mathrm{m}\times \left(-1\right) \\ =0.9 \mathrm{J} \end{gathered}$$

(c)      

Work done on the book by the normal force from the tabletop

$$\begin{gathered} {\mathrm{W}}_{\mathrm{N}}=\mathrm{Fs} \cos 180° \\ =0 \end{gathered}$$ 

(d)

Work done on the book by the gravity will be

${\mathrm{W}}_{\mathrm{g}}=\mathrm{Fs} \cos 90°$

       = 0

Since F and s are perpendicular to each other.

$$\begin{gathered} {\mathrm{W}}_{\mathrm{g}}=\mathrm{Fs} \cos \left(-90°\right) \\ =0 \end{gathered}$$ 

(e)

Net work done on the book is equal to the sum of the individual work

$$\begin{gathered} {\mathrm{W}}_{\mathrm{net}}={\mathrm{W}}_{\mathrm{push}}+{\mathrm{W}}_{\mathrm{f}}+{\mathrm{W}}_{\mathrm{N}}+{\mathrm{W}}_{\mathrm{g}} \\ =3.6 \mathrm{J}-0.9 \mathrm{J}+0 \mathrm{J}+0 \mathrm{J} \\ =2.7 \mathrm{J} \end{gathered}$$

Thus, the work done by various forces on the book is ${\mathrm{W}}_{\mathrm{push}}=3.6 \mathrm{J},{\mathrm{W}}_{\mathrm{f}}=-0.9 \mathrm{J},{\mathrm{W}}_{\mathrm{N}}=0 \mathrm{J},{\mathrm{W}}_{\mathrm{net}}=2.7 \mathrm{J}$