91Ó°ÊÓ

Force constant = 76.0 N

Block = 2.00 kg

Constant horizontal force = 54.0 N

The work-energy theorem when applied on the block is,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{tot}}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ {\mathrm{W}}_{\mathrm{F}}+{\mathrm{W}}_{\mathrm{spring}}=\frac{1}{2}\mathrm{mv}{\text{'}}^2-\frac{1}{2}{\mathrm{mv}}^2 \\ {\mathrm{F}}_{\mathrm{x}}+\left(-\frac{1}{2}{\mathrm{kx}}^2\right)=\frac{1}{2}\mathrm{mv}{\text{'}}^2-\frac{1}{2}{\mathrm{mv}}^2 \end{gathered}$$

Substitute $9.8\mathrm{m}/{\mathrm{s}}^2$for g , 2.00 kg for m , 76.0 N for k , 54.0 N for F , 0.400 m for x , and 0 m/s for v.

$$\begin{gathered} \left\{\left(54.0\mathrm{N}\right)\left(0.400\mathrm{m}\right)-\frac{1}{2}\left(76.0\mathrm{N}/\mathrm{m}\right){\left(0.400\mathrm{m}\right)}^2\right\} \\ \mathrm{v}\text{'}=\frac{\sqrt{\left\{\left(54.0\mathrm{N}\right)\left(0.400\mathrm{m}\right)-{\displaystyle \frac{1}{2}}\left(76.0\mathrm{N}/\mathrm{m}\right){\left(0.400\mathrm{m}\right)}^2\right\}}}{{\displaystyle \frac{1}{2}}\left(2.00\mathrm{kg}\right)} \\ \mathrm{v}\text{'}=3.94\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the block is 3.94 m/s

With the assumption that a positive direction is away from the post, Newton’s second law gives,

$$\begin{gathered} \mathrm{F}-{\mathrm{F}}_{\mathrm{spring}}=\mathrm{ma} \\ \mathrm{F}-\mathrm{kx}=\mathrm{ma} \end{gathered}$$

Here, a is the acceleration of the block.

Substitute $9.8\mathrm{m}/{\mathrm{s}}^2\mathrm{for}\mathrm{g},2.00\mathrm{kg}\mathrm{for}\mathrm{m},76.0\mathrm{N}\mathrm{for}\mathrm{k},54.0\mathrm{N}\mathrm{for}\mathrm{F},0.400\mathrm{m}\mathrm{for}\mathrm{x}$

$$\begin{gathered} \mathrm{a}=\frac{54.0\mathrm{N}-\left(76.09\mathrm{N}/\mathrm{m}\right)×\left(0.400\mathrm{m}\right)=\left(2.00\mathrm{kg}\right)\mathrm{a}}{\left(2.00\mathrm{kg}\right)} \\ \mathrm{a}=11.8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The positive sign indicates that the direction of acceleration is away from the post.

Hence, the direction of the acceleration is away from the post, and the magnitude of the acceleration is $11.8\mathrm{m}/{\mathrm{s}}^2$.