One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 2.00 - kg block of frictionless ice is attached to the other end and rests on the floor. The spring is initially neither stretched nor compressed. A constant horizontal force of 54.0 N is then applied to the block, in the direction away from the post. (a) What is the speed of the block when the spring is stretched 0.400 m ? (b) At that instant, what are the magnitude and direction of the acceleration of the block?

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

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

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 $11.8\mathrm{m}/{\mathrm{s}}^2$.