Chapter 6, Problem 118P

(a)

To determine

The maximum compression of the spring.

Answer to Problem 118P

The maximum compression distance of the spring is $26\text{cm}$.

Explanation of Solution

The figure 1 shows the forces acting on the block.

Write the equation for mechanical energy of the system

$K_i+U_{i,g}+U_{i,e}+W=K_f+U_{f,g}+U_{f,e}$ (I)

Here, $K_i$ is the initial kinetic energy of the block, $K_f$ is the final kinetic energy of the block, $U_{i,g}$ is the initial potential energy due to gravity, $U_{f,g}$ is the final potential energy due to gravity, $U_{i,e}$ is the initial elastic potential energy of the spring, $U_{f,e}$ is the final elastic potential energy of the spring, and $W$ is the work done due to frictional force.

Write the equation for Newton’s second law of motion of the block with perpendicular to the plane in y direction.

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ N-mg\cos \theta =0\end{array}$ (II)

Here, $N$ is the normal force acting on the block, $m$ is the mass of the block, $g$ is the gravitational acceleration, and $\theta$ is the angle between the weight and the normal to the incline.

Rewrite the equation (II) for normal force.

$N=mg\cos \theta$ (III)

Write the equation for the frictional force.

$f=\mu N$ (IV)

Here, $f$ is the frictional force and $\mu$ is the coefficient of friction.

Substitute equation (III) in equation (IV).

$f=\mu mg\cos \theta$ (V)

Replace $1/2k{x}_f^2$ for $U_{f,e}$, $mg{y}_i$ for $U_{i,g}$, and $f_{k}d\cos 180°$ for $W$ in equation (I).

$\begin{array}{c}0+mg{y}_i+0+f_{k}d\cos 180°=0+0+\frac{1}{2}k{x}_f^2\\ mg{y}_i-f_{k}d=\frac{1}{2}k{x}_f^2\end{array}$ (VI)

Here, $y_i$ is the initial sliding distance of the block along the horizontal line, $f_k$ is the kinetic friction, $d$ is the sliding distance of the block, $k$ is the spring constant, and $x_f$ is the compression distance of the spring.

Substitute equation (V) in equation (VI).

$mgd\sin \theta -\mu mgd\cos \theta =\frac{1}{2}k{x}_f^2$ (VII)

Conclusion:

Solve the equation (VII) for $x_f$.

$\begin{array}{c}mgd\left(\sin \theta -\mu \cos \theta \right)=\frac{1}{2}k{x}_f^2\\ x_f^2=\frac{2mgd}{k}\left(\sin \theta -\mu \cos \theta \right)\\ x_f=\sqrt{\frac{2mgd}{k}\left(\sin \theta -\mu \cos \theta \right)}\end{array}$

Substitute $0.50\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, $85\text{cm}$ for $d$, $35\text{N/m}$ for $k$, $0.25$ for $\mu$, and $30.0°$ for $\theta$ in above equation.

$\begin{array}{c}{x}_f=\sqrt{\frac{2\left(0.50\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\left(85\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)}{\left(35\text{N/m}\right)}\left(\sin 30.0°-\left(0.25\right)\cos 30.0°\right)}\\ =0.2595\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =25.95\text{cm}\\ =26\text{cm}\end{array}$

Therefore, the maximum compression distance of the spring is $26\text{cm}$.

(b)

To determine

The distance of the block inclined before coming to rest.

Answer to Problem 118P

The distance of the block inclined before coming to rest is $34\text{cm}$.

Explanation of Solution

Write the equation (VII) from part (a) for the conservation of  work –energy theorem.

$\begin{array}{c}mgd\sin \theta -\mu mgd\cos \theta =mg{d}^{\prime }\sin \theta +\mu mg{d}^{\prime }\cos \theta \\ mgd\left(\sin \theta -\mu \cos \theta \right)=mg{d}^{\prime }\left(\sin \theta +\mu \cos \theta \right)\\ d\left(\sin \theta -\mu \cos \theta \right)=d^{\prime }\left(\sin \theta +\mu \cos \theta \right)\\ d^{\prime }=\frac{d\left(\sin \theta -\mu \cos \theta \right)}{\left(\sin \theta +\mu \cos \theta \right)}\end{array}$ (VIII)

Conclusion:

Substitute $85\text{cm}$ for $d$, $0.25$ for $\mu$, and $30.0°$ for $\theta$ in equation (VIII).

$\begin{array}{c}{d}^{\prime }=\frac{\left(85\text{cm}\right)\left(\sin 30.0°-\left(0.25\right)\cos 30.0°\right)}{\left(\sin 30.0°+\left(0.25\right)\cos 30.0°\right)}\\ =\left(85\text{cm}\right)\frac{0.5-0.217}{0.5+0.217}\\ =\left(85\text{cm}\right)\frac{0.283}{0.717}\\ =34\text{cm}\end{array}$

Therefore, the distance of the block inclined before coming to rest is $34\text{cm}$.