Chapter 6, Problem 142P

(a)

To determine

The speed of the block along the incline.

Answer to Problem 142P

The speed of the block along the incline is $\underset{\_}{2\text{m}/\text{s}}$.

Explanation of Solution

Figure 1 represents the motion of the block along the incline.

Consider the block is at a height $h$ initially with the angle of inclination $\theta$.

Apply conservation of energy to Figure 1.

    $mgh=mg{h}_2+\frac{1}{2}m{v}^2+f_{\text{k}}l$

Here, $m$ is the mass of the block, $g$ is the acceleration due to gravity, $f_{\text{k}}$ is the frictional force, $l$ is the distance moved along the incline, $h$ is the initial height of the block, $v$ is the speed of the block and $h_2$ is the height of the block at the intermediate stage of motion.

Rearrange the above equation.

    $mg\left(h-h_2\right)=\frac{1}{2}m{v}^2+f_{\text{k}}l$        (I)

From Figure 1 $h-h_2=l\sin \theta$ and the frictional force $f_{\text{k}}={\mu }_{\text{k}}mg\cos \theta$.

Apply the above condition in equation (I). and rearrange to find $v$.

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

Conclusion:

Substitute $9.8\text{m}/{\text{s}}^2$ for $g$, $\text{30}\text{cm}$ for $l$, $0.38$ for ${\mu }_{\text{k}}$, and $60°$ for $\theta$ in equation (II), to find $v$.

    $\begin{array}{c}v=\sqrt{2\left(9.8\text{m}/{\text{s}}^2\right)\left(\text{30}\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)\left(\sin 60°-{\mu }_{\text{k}}\cos 60°\right)}\\ =1.993\text{m}/\text{s}\\ \approx 2\text{m}/\text{s}\end{array}$

Therefore, the speed of the block along the incline is $\underset{\_}{2\text{m}/\text{s}}$.

(b)

To determine

The speed of the block along the incline using newton’s second law of motion, to check which method is easier to calculate the speed.

Answer to Problem 142P

The speed of the block along the incline using newton’s second law of motion is $\underset{\_}{2\text{m}/\text{s}}$ , calculating the speed using newton’s second law of motion is much easier.

Explanation of Solution

Figure 2 represents the free body diagram of the block.

Write the net force acting on the block in the vertical direction.

    $\begin{array}{c}N-mg\cos \theta =0\\ N=mg\cos \theta \end{array}$        (III)

Here, $N$ is the normal force.

Write the net force acting on the block in the horizontal direction.

    $F=mg\sin \theta -{\mu }_{\text{k}}N$        (IV)

Here ${\mu }_{\text{k}}$ is the coefficient of kinetic friction.

Use equation (III) in (IV).

    $F=mg\sin \theta -{\mu }_{\text{k}}mg\cos \theta$        (V)

Write the expression for acceleration of the block.

    $a=\frac{F}{m}$

Here, $a$ is the acceleration of the block, $F$ is the force experienced by the block along the incline, and $m$ is the mass of the block.

Use equation (V) in the above equation.

    $\begin{array}{c}a=\frac{mg\sin \theta -{\mu }_{\text{k}}mg\cos \theta }{m}\\ =g\sin \theta -{\mu }_{\text{k}}g\cos \theta \end{array}$        (VI)

Write the kinematic equation.

    $v^2-u^2=2as$

Here, $v$ is the final speed of the block, $u$ is the initial speed of the speed, and $s$ is the distance travelled along the incline.

Since the initial speed of the block is zero, the above equation is reduced to

    $\begin{array}{c}{v}^2=2as\\ v=\sqrt{2as}\end{array}$        (VII)

Conclusion:

Substitute $9.8\text{m}/{\text{s}}^2$ for $g$, $0.38$ for ${\mu }_{\text{k}}$, and $60°$ for $\theta$ in equation (VI), to find $a$.

    $\begin{array}{c}a=\left(9.8\text{m}/{\text{s}}^2\right)\sin 60°-\left(0.38\right)\left(9.8\text{m}/{\text{s}}^2\right)\cos 60°\\ =6.625\text{m}/{\text{s}}^2\end{array}$

Substitute $6.625\text{m}/{\text{s}}^2$ for $a$, and $\text{30}\text{cm}$ for $s$ in equation (VII), to find $v$.

    $\begin{array}{c}v=\sqrt{2\left(6.625\text{m}/{\text{s}}^2\right)\left(\text{30}\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}\\ =1.993\text{m}/\text{s}\\ \approx 2\text{m}/\text{s}\end{array}$

Therefore, the speed of the block along the incline using newton’s second law of motion is $\underset{\_}{2\text{m}/\text{s}}$ , calculating the speed using newton’s second law of motion is much easier.