91Ó°ÊÓ

1. Distance traveled is, d=2.00 m.
2. Initial kinetic energy, ${\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{initial}}=3.00\mathrm{J}$.

The problem deals with the work-energy theorem. It states that the net work done by the forces on an object equals the change in its kinetic energy. We can use the work-energy theorem to find the final kinetic energy of the can.

Fromthegraph, we can conclude that,

At x=2.00 m value for work done is, $\mathrm{W}=6.0\mathrm{J}$.

The relation between the work done W by the force F while moving the object with a distance d, can be written as,

$\mathrm{W}=\mathrm{Fd}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}6.0\mathrm{N}.\mathrm{m}& =& \mathrm{F}×2.00\mathrm{m}\\ \mathrm{F}& =& \frac{6.0\mathrm{N}.\mathrm{m}}{2.00\mathrm{m}}\\ \mathrm{F}& =& 3.0\mathrm{N}\end{array}$

Thus, the magnitude of the horizontal force acting on the can is 3.00 N.

According to the energy-work theorem, the change in the kinetic energy is equal to the work done.

We can write it as,

$\begin{array}{rcl}\mathrm{W}& =& ∆\mathrm{KE}\\ \mathrm{W}& =& {\left(\mathrm{KE}\right)}_{\mathrm{Final}}-{\left(\mathrm{KE}\right)}_{\mathrm{Initial}}\\ {\left(\mathrm{KE}\right)}_{\mathrm{Final}}& =& \mathrm{W}+{\left(\mathrm{KE}\right)}_{\mathrm{Initial}}\end{array}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} {\left(\mathrm{KE}\right)}_{\mathrm{Final}}=6.00\mathrm{J}+3.00\mathrm{J} \\ {\left(\mathrm{KE}\right)}_{\mathrm{Final}}=9.00\mathrm{J} \end{gathered}$$

Thus, the end kinetic energy of the can at x=2.00 m is 9.00 J.