Chapter 6, Problem 36P

(a)

To determine

The speed of the brick when it reaches edge of the roof.

Answer to Problem 36P

The speed of the brick when it reaches edge of the roof is $4.43\text{ m/s}$ .

Explanation of Solution

Write the equation for the change in height.

$\Delta y=l\sin \theta$ (I)

Here, $\Delta y$ is the change in height, $l$ is the distance to the edge of the roof, and $\theta$ is the angle of inclination of the roof

Write the principle of conservation of energy.

$K_i+U_i=K_f+U_f$

Here, $K_i$ is the initial kinetic energy of the brick, $K_f$ is the final kinetic energy of the brick, $U_i$ is the initial potential energy of the brick and $U_f$ is the final potential energy of the brick

Since the brick starts from rest, its initial kinetic energy is zero and also its potential energy at the edge of the roof is zero.

Substitute $0$ for $K_i$ and $U_f$ in the above equation.

$\begin{array}{c}0+U_i=K_f+0\\ U_i=K_f\end{array}$ (II)

Write the equation for $U_i$ .

$U_i=mg\Delta y$ (III)

Here, $m$ is the mass of the brick and $g$ is the acceleration due to gravity

Write the expression for $K_f$ .

$K_f=\frac{1}{2}m{v}^2$ (IV)

Here, $v$ is the constant speed of the brick

Put equations (III) and (IV) in equation (II).

$\begin{array}{c}mg\Delta y=\frac{1}{2}m{v}^2\\ v^2=2g\Delta y\\ v=\sqrt{2g\Delta y}\end{array}$ (V)

Conclusion:

Given that the distance to the edge of the roof is $2.00\text{ m}$ and the angle of inclination of the roof

 is $30.0°$ . The value of acceleration due to gravity is $9.80{\text{ m/s}}^2$ .

Substitute $2.00\text{ m}$ for $l$ and $30.0°$ for $\theta$ in equation (I) to find $\Delta y$ .

$\begin{array}{c}\Delta y=\left(2.00\text{ m}\right)\sin 30.0°\\ =1.00\text{ m}\end{array}$

Substitute $9.80{\text{ m/s}}^2$ for $g$ and $1.00\text{ m}$ for $\Delta y$ in equation (V) to find $v$ .

$\begin{array}{c}v=\sqrt{2\left(9.80{\text{ m/s}}^2\right)\left(1.00\text{ m}\right)}\\ =4.43\text{ m/s}\end{array}$

Therefore, the speed of the brick when it reaches edge of the roof is $4.43\text{ m/s}$ .

(b)

To determine

The speed of the brick when it reaches edge of the roof if the coefficient of kinetic friction is $0.10$ .

Answer to Problem 36P

The speed of the brick when it reaches edge of the roof with coefficient of kinetic friction $0.10$ is $4.03\text{ m/s}$ .

Explanation of Solution

Write the equation for the work done by the friction force on the brick.

$\begin{array}{c}{W}_f=Fl\cos 180°\\ =-Fl\end{array}$ (VI)

Here, $W_f$ is the work done by the friction force on the brick and $f$ is the magnitude of force of friction

Write the equation for $F$ .

$F=\mu mg\cos \theta$

Here, $\mu$ is the coefficient of kinetic friction

Put the above equation in equation (VI).

$W_f=-\mu mgl\cos \theta$

This work is stored as the potential energy.

Write the equation for $U_i$ .

$U_i=mg\Delta y-\mu mgl\cos \theta$

Put equation (I) in the above equation.

$\begin{array}{c}{U}_i=mgl\sin \theta -\mu mgl\cos \theta \\ =mgl\left(\sin \theta -\mu \cos \theta \right)\end{array}$ (VII)

Put equations (IV) and (VII) in equation (II).

$\begin{array}{c}\frac{1}{2}m{v}^2=mgl\left(\sin \theta -\mu \cos \theta \right)\\ v^2=2gl\left(\sin \theta -\mu \cos \theta \right)\\ v=\sqrt{2gl\left(\sin \theta -\mu \cos \theta \right)}\end{array}$ (VIII)

Conclusion:

Substitute $9.80{\text{ m/s}}^2$ for $g$ , $2.00\text{ m}$ for $l$ , $30.0°$ for $\theta$ and $0.10$ for $\mu$ in equation (VIII) to find $v$ .

$\begin{array}{c}v=\sqrt{2\left(9.80{\text{ m/s}}^2\right)\left(2.00\text{ m}\right)\left(\sin 30.0°-0.10\cos 30.0°\right)}\\ =4.03\text{ m/s}\end{array}$

Therefore, the speed of the brick when it reaches edge of the roof with coefficient of kinetic friction $0.10$ is $4.03\text{ m/s}$ .