Figure 7 -42 shows a cold package of hot dogs sliding rightward across a frictionless floor through a distance $\mathit{d}\mathbf{}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$while three forces act on the package. Two of them are horizontal and have the magnitudes ${\mathit{F}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}$and${\mathit{F}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}$; the third is angled down by$\mathbf{\theta }\mathbf{}\mathbf{=}\mathbf{60}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$ and has the magnitude${\mathit{F}}_{\mathbf{3}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}$. (a) For the 20.0 cm displacement, what is the net work done on the package by the three applied forces, the gravitational force on the package, and the normal force on the package? (b) If the package has a mass of 2.0 Kg and an initial kinetic energy of 0, what is its speed at the end of the displacement?

It is given that,

Distance$d=20.0\mathrm{cm}=0.200\mathrm{m}$.

Two forces are acting horizontal, $F_1=5.00\mathrm{N}$and $F_2=1.00\mathrm{N}$.

One force is acting at angle $\theta =60.0°$ down, $F_3=4.00\mathrm{N}$.

Displacement is$20.0\mathrm{cm}=0.200\mathrm{m}$

Mass of the package is$m=2.0\mathrm{Kg}$.

Initial kinetic energy is 0

First, find the x component of the force ${\mathit{F}}_{\mathbf{3}}$.By using this value; find the total force $\mathit{F}$ acting on the package. By using the values of $\mathit{F}$ and d, find the value of the work done Won the package by the given three applied forces, and using this value of W , find the package speed Vat the end of the displacement.

Formulae:

The work done is given by,

$W=Fd$

The speed is,

$v=\sqrt{\frac{2W}{m}}$

where, Fis force, d is displacement, Wis the work done, vis velocity and m is mass.

Thecomponent of force${\mathrm{F}}_3$is,

$F_{3x}=(4.00\mathrm{N})\cos \left(60\right)$

The total force applied on the package is,

$$\begin{gathered} F=F_1-F_2+F_3 \\ F=5.00-1.00+(4.00)\cos \left(60\right) \\ F=6.00\mathrm{N} \end{gathered}$$

Therefore, the work done is given by,

$$\begin{gathered} W=F.d \\ W=6.00\times 0.200 \\ W=1.20\mathrm{J} \end{gathered}$$

Hence, the work done on the package by the three given applied forces is $W=1.20\mathrm{J}$

Fromthework energy theorem, the work done is,

$$\begin{gathered} W=\frac{1}{2}m{v}^2 \\ v=\sqrt{\frac{2W}{m}} \\ v=\sqrt{\frac{2\times 1.20}{2.0}} \\ v=1.10\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the package speed at the end of the displacement is $v=1.10\mathrm{m}/\mathrm{s}$