The only force acting on abody as the body moves along an x axis varies as shown in Fig.. The scale of the figure’s vertical axis is set by. The velocity of the body atis. (a) What is the kinetic energy of the body at? (b) At what value of x will the body have a kinetic energy of? (c) What is the maximum kinetic energy of the body betweenand?

Mass of the object is$m=2.0\mathrm{kg}$

Force is$F_s=4.0\mathrm{N}$

Initial velocity is$v_i=4.0\mathrm{m}/\mathrm{s}$

This problem deals with the work-energy theorem which states that the net work done by the forces applied on object is equal to the change in its kinetic energy. Use the work-energy theorem to solve this problem. To initiate, the work done can be found by calculating the area under the graph. Further, kinetic energies of the body at the given positions will be calculated. Using an equation for work in terms of force and displacement to the distance for the given kinetic energy can be computed. For maximum kinetic energy, use positive work.

Formulae are as follow:

$$\begin{gathered} W=\frac{1}{2}m{v^2}_f-\frac{1}{2}m{v^2}_i \\ W=F.d \end{gathered}$$

where, Fis the force, d is the displacement, Wis the work done, $v_i,v_f$are initial and final velocities and m is the mass.

Firstly, find the work done from the graph by calculating the area under the graph.

For the work done for$x=0$to$x=1$,

$W_1=\frac{1}{2}\left(1\right)(4.0)=2\mathrm{J}$

For the work done for$x=1$to$x=2$,

$W_2=\frac{1}{2}\left(1\right)(-4.0)=-2\mathrm{J}$

For the work done for$x=2$to$x=3$,

$W_3=(-4)\left(1\right)=-4\mathrm{J}$

So, total work done is,

$$\begin{gathered} W=W_1+W_2+W_3 \\ W=2+(-2)+(-4)=-4\mathrm{J} \\ W=-4\mathrm{J} \end{gathered}$$

Work done is the change in kinetic energy,

localid="1657179960656" $$\begin{gathered} W=\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}-\mathrm{K}.{\mathrm{E}}_{\mathrm{I}} \\ \mathrm{K}.{\mathrm{E}}_f=\frac{1}{2}m{v^2}_i+W \\ \mathrm{K}.{\mathrm{E}}_f=\frac{1}{2}(2.0){(4.0)}^2-4\mathrm{K}.{\mathrm{E}}_f=12\mathrm{J} \end{gathered}$$

Hence, the kinetic energy of the body at $\times =3.0\mathrm{m}$is$\mathrm{K}.\mathrm{E}.=12\mathrm{J}$.

Kinetic energy at $\times =3\mathrm{mis}12\mathrm{J}$. The given kinetic energy is 8.0J.Therefore, the distance will be$(\times -3),$

$$\begin{gathered} F_s\times (x-3)=\mathrm{K}.{\mathrm{E}}_{\mathrm{x}}-\mathrm{K}.{\mathrm{E}}_3 \\ {F}_s\times (x-3)=8.0-12 \end{gathered}$$

$$\begin{gathered} -4\times (x-3)=-4 \\ x=4\mathrm{m} \end{gathered}$$

Hence, the value of x where kinetic energy of body will be $8.0\mathrm{J}$is$x=4\mathrm{m}$ .

Kinetic energy increases when work is positive, so at$x=1\mathrm{m}$the work is positive.

role="math" localid="1657179818473" $$\begin{gathered} W_1=\mathrm{K}.\mathrm{E}.\text{'}-\mathrm{K}.{\mathrm{E}}_{\mathrm{i}} \\ 2=\mathrm{K}.\mathrm{E}.\text{'}-16 \\ \mathrm{K}.\mathrm{E}.\text{'}=18\mathrm{J} \end{gathered}$$

Hence, the maximum kinetic energy of the body between $x=0$and $x=5.0\mathrm{m}$is$\mathrm{K}.\mathrm{E}.\text{'}=18\mathrm{J}$ .