One end of a horizontal spring with force constant 130N/m is attached to a vertical wall. A 4kg block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is ${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}$. You apply a constant force $\overrightarrow{\mathbf{F}}$ to the block. $\overrightarrow{\mathbf{F}}$ has magnitude $\mathbf{F}\mathbf{=}\mathbf{82}\mathbf{}\mathbf{N}$ and is directed toward the wall. At the instant that the spring is compressed 80.0 cm, what are (a) the speed of the block, and (b) the magnitude and direction of the block’s acceleration?

The given data can be listed below,

• The force constant of the spring is, $k=130\mathrm{N}/\mathrm{m}$.
• The mass of the block is, $m=4\mathrm{kg}$.
• The kinetic friction is, ${\mu }_k=0.40$
• The magnitude of force is, $F=82\mathrm{N}$.
• The compression of the spring is, $x=80\mathrm{cm}$

The coefficient of friction is the ratio of frictional force and normal reaction force acting on the body.

The free body diagram is given by,

The normal force on the block is given by,

$N=mg$

Here, m is the mass of the block and g is the acceleration due to gravity.

Substitute all the values in the above,

$$\begin{gathered} N=4\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =39.2\mathrm{N} \end{gathered}$$

The friction force on the block is given by,

$F_k={\mu }_{k}N$

Here, ${\mu }_k$ is the coefficient of friction, and Nis the normal force.

Substitute all the values in the above,

$$\begin{gathered} F=0.40\left(39.2\right) \\ =15.68\mathrm{N} \end{gathered}$$

The work done is given by,

$W_F=Fx\cos 0°$

Here, F is the constant external force, and x is the compressed distance.

Substitute all the values in the above,

$$\begin{gathered} W_F=\left(82\mathrm{N}\right)\left(0.80\mathrm{m}\right)\left(1\right) \\ =65.6\mathrm{J} \end{gathered}$$

The work done on the spring is given by,

$W_s=\frac{1}{2}k{x}^2$

Here, k is the force constant, and x is the length of compression.

Substitute all the values in the above,

role="math" localid="1665143890891" $$\begin{gathered} W_s=\frac{1}{2}\left(130\mathrm{N}/\mathrm{m}\right){\left(0.80\mathrm{m}\right)}^2 \\ =-41.6\mathrm{J} \end{gathered}$$

The work done due to friction force is given by,

$W_{fk}=F_{k}x\cos 180°$

Substitute all the values in the above,

$$\begin{gathered} W_{fk}=15.60\mathrm{N}\left(0.80\mathrm{m}\right)\cos 180° \\ =-12.54\mathrm{J} \end{gathered}$$

The net work done is given by,

$W=W_F+W_s+W_{fk}$

Substitute all the values in the above,

$$\begin{gathered} W=\left(65.6-41.6-12.54\right)\mathrm{J} \\ =11.54\mathrm{J} \end{gathered}$$

Also, the net work done of the block is difference between the kinetic energies which can be given by,

$W=\frac{1}{2}m\left(v_1^2-v_0^2\right)$

Here, m is the mass of the block, $v_1$ is the final velocity of the block and $v_0$ is the initial velocity of the block whose value is zero.

Substitute all the values in the above, the final velocity of the

$$\begin{gathered} \mathrm{w}=\frac{1}{2}\left(4\mathrm{kg}\right)\left({\mathrm{v}}_1^2-0\right) \\ {\mathrm{v}}_1=\sqrt{2\mathrm{W}} \\ =\sqrt{2\times 11.45} \\ =2.39\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the block is $2.39\mathrm{m}/\mathrm{s}$

The acceleration of the block is given by,

$$\begin{gathered} \sum _{}{F}_x=m{a}_x \\ =f_k+S-F \\ {a}_x=\frac{f_k+S-F}{m} \end{gathered}$$

Substitute all the values in the above,

$$\begin{gathered} a_x=\frac{15.68\mathrm{N}+\left(130\mathrm{N}/\mathrm{m}\right)\left(0.80\mathrm{m}\right)-82\mathrm{N}}{4\mathrm{kg}} \\ =\frac{37.68}{4} \\ =9.42\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration of the block is in right direction and its value is $9.42\mathrm{m}/{\mathrm{s}}^2$.