Question

A block of mass $\mathit{m}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathit{k}\mathit{g}$is pushed a distance $\mathit{d}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{2}\mathit{m}$along a frictionless, horizontal table by a constant applied force of magnitude $\mathit{F}\mathbf{=}\mathbf{16}\mathit{N}$ directed at an angle $\mathit{\theta }\mathbf{=}\mathbf{25}\mathbf{°}$ below the horizontal as shown in Figure P7.5. Determine the work

done on the block by

(a) the applied force,

(b) the normal force exerted by the table,

(c) the gravitational force, and

(d) the net force on the block.

Short answer

(a) The work done on the block by the applied force is$\mathbf{31}\mathbf{.}\mathbf{9}\mathit{J}$ .

(b) The work done on the block by the normal force is .$\mathbf{0}\mathit{J}$

(c) The work done on the block by the gravitational force is .$\mathbf{0}\mathit{J}$

(d) The word done on the block by the net force is .$\mathbf{31}\mathbf{.}\mathbf{9}\mathit{J}$

Step-by-step solution

Given data

The mass of the block is

$m=2.5\text{\hspace{0.17em}kg}$

The horizontal displacement of the block is

$d=2.2\text{\hspace{0.17em}m}$

The applied force is

$F=16\text{\hspace{0.17em}N}$

The angle made by the applied force with the horizontal is

$\theta =25°$

Work done

The work done by a force $F$ causing displacement $d$ making an angle $\theta$ with it is

$\mathit{W}\mathbf{=}\mathit{F}\mathit{d}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$ .....(I)

Determining the work done by the applied force

The force$F=16\text{\hspace{0.17em}N}$makes and angle $\theta =25°$ with the direction of displacement. Thus from equation (I), the work done by it is

$\begin{array}{c}W=16\text{\hspace{0.17em}N}\times 2.2\text{\hspace{0.17em}m}\times \cos 25°\\ =31.9\cdot (1\text{\hspace{0.17em}N}\cdot \text{m}\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}N}\cdot \text{m}})\\ =31.9\text{\hspace{0.17em}J}\end{array}$

The required work done is.$31.9\text{\hspace{0.17em}J}$

Determining the work done by the normal force

The normal force $F_n$ by the table on the block is directed vertically upward and the displacement of the book is horizontal. Thus the angle between the normal force and the displacement is $90°$. From equation (I) the work done by the normal force is

$\begin{array}{c}W=F_n\times 2.2\text{\hspace{0.17em}m}\times \cos 90°\\ =0\cdot (1\text{\hspace{0.17em}N}\cdot \text{m}\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}N}\cdot \text{m}})\\ =0\text{\hspace{0.17em}J}\end{array}$

The required work done is .$0\text{\hspace{0.17em}J}$

Determining the work done by the gravitational force

The gravitational force $F_g$by the table on the block is directed vertically downward and the displacement of the book is horizontal. Thus the angle between the gravitational force and the displacement is.$90°$From equation (I) the work done by the gravitational force is

$\begin{array}{c}W=F_g\times 2.2\text{\hspace{0.17em}m}\times \cos 90°\\ =0\cdot (1\text{\hspace{0.17em}N}\cdot \text{m}\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}N}\cdot \text{m}})\\ =0\text{\hspace{0.17em}J}\end{array}$

The required work done is .$0\text{\hspace{0.17em}J}$

Determining the work done by the net force on the block

Since the block has no vertical displacement, the net vertical force on it is zero. Thus the gravitational force and normal force cancel each other. Thus the net force on the block is the applied force .$F=16\text{\hspace{0.17em}N}$ The work done by this applied force, as calculated before is .$31.9\text{\hspace{0.17em}J}$