91Ó°ÊÓ

We can find the velocity and value ofx at$\mathit{v}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{ }\mathbf{\text{m/s}}$of the body by equating thework done obtained fromtheintegral of the x-component of the applied force and change in the K.E according to the work-energy theorem.

Given:

1. x-component of the force acting on the body is$F=-6x \text{N}.$
2. Mass of the body is$m=2 \text{kg}$
3. The velocity at $x=3.0 \text{m}$is$8.0 \text{m/s}$

Formula:

$W=\stackrel{→}{F}.\stackrel{→}{d}$

$W=\mathrm{Δ}K.E=\frac{1}{2}m{\left(\mathrm{Δ}v\right)}^2$

Work done by the x-component of the force is,

$W={∫}_{x_i}^{x_f}{F}_x$

Substitute the value of$F_x$in the above equation.

$$\begin{gathered} W={∫}_{x_i}^{x_f}-6xdx \\ W=-3\left(x_f^2-x_i^2\right) \end{gathered}$$

Body moves from $\text{x}=3 m$to $\text{x}=4 m$.

$$\begin{gathered} \text{W}=-3\left({\left(4 \text{m}\right)}^2-{\left(3 \text{m}\right)}^2\right) \\ W=-21 \text{J} \end{gathered}$$

According to the work-energy theorem,

$\begin{array}{rcl}W& =& K.E\\ & =& \frac{1}{2}m\left(v_f^2-v_i^2\right)\\ & & \end{array}$

$$\begin{gathered} \frac{1}{2}\text{m}\left({\text{v}}_{\text{f}}^{\text{2}}-{\text{v}}_{\text{i}}^{\text{2}}\right)=-21 \text{J} \\ \left(v_f^2-v_i^2\right)=\frac{2}{m}\left(-21 \text{J}\right) \\ {v}_f^2=v_i^2+\frac{2}{m}\left(-21 \text{J}\right) \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} v_f^2={\left(8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2+\frac{2}{2 \text{kg}}\left(-21 \text{J}\right) \\ {v}_f=6.557\raisebox{1ex}{${\displaystyle m}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle s}$}\right. \\ {v}_f≃6.6 \raisebox{1ex}{${\displaystyle m}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle s}$}\right. \end{gathered}$$

Therefore, the velocity of body is,$6.6\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

From part a, we can write,

$$\begin{gathered} -3\left(x_f^2-x_i^2\right)=\frac{1}{2}m\left(v_f^2-v_i^2\right) \\ {x}_f^2=\frac{m}{2\left(-3\right)}\left(v_f^2-v_i^2\right)+x_i^2 \\ {x}_f=\sqrt{\frac{m}{\left(-6\right)}\left(v_f^2-v_i^2\right)+x_i^2} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} x_f=\sqrt{\frac{2 \text{kg}\left({\left({\displaystyle 5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\right)}^{\text{2}}\text{-}{\left(8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\right)}^{\text{2}}\right)\text{+}{\left(3m\right)}^{\text{2}}}{\left(-6\right)}} \\ =\sqrt{\left(22 {\text{m}}^2\right)} \\ {x}_f=4.69 \text{m} \end{gathered}$$

Therefore, the positive value of at which the body has a velocity of $5.0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$is$4.7 \text{m}$.