91Ó°ÊÓ

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, ${\mathrm{F}}_{\mathrm{f}}=0.6\mathrm{N}$

Work done on a particle by a constant forceduring a linear displacementis given by

$$\begin{gathered} \mathbf{W}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{s}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{¹ó³¦´Ç²õÏ•} \end{gathered}$$ …â¶Ä¦â¶Ä¦â¶Ä¦..(1)

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

$\mathrm{ϕ}=180°$.

(a)

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

$\mathrm{W}=\mathrm{¹ó²õ³¦´Ç²õÏ•}$

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

Therefore, W = F s cos0

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

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

(b)

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

${\mathrm{W}}_{\mathrm{f}}=\mathrm{Fs}\mathrm{³¦´Ç²õÏ•}$

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

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

role="math" localid="1663748170934" $$\begin{gathered} \mathrm{For}\mathrm{F}=0.6\mathrm{N},\mathrm{s}=1.5\mathrm{m} \\ \mathrm{W}=\mathrm{Fscos}180° \\ =0.6\mathrm{N}×1.5\mathrm{m}×\left(-1\right) \\ =0.9\mathrm{J} \end{gathered}$$

(c)

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

role="math" localid="1663748300554" $$\begin{gathered} {\mathrm{W}}_{\mathrm{N}}=\mathrm{Fs}\cos 180° \\ =0 \end{gathered}$$

(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${\mathrm{W}}_{\mathrm{push}}=3.6\mathrm{J},{\mathrm{W}}_{\mathrm{f}}=-0.9\mathrm{J},{\mathrm{W}}_{\mathrm{N}}=0\mathrm{J},{\mathrm{W}}_{\mathrm{net}}=2.7\mathrm{J}$