91Ó°ÊÓ

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 $\mathrm{θ}=60.0°$ 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,

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

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,

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

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