Chapter 4, Problem 83P

(a)

To determine

The initial speed of the block.

Answer to Problem 83P

The initial speed of the block is $\underset{\_}{9.4\text{m}/\text{s}}$.

Explanation of Solution

Figure 1 represents that a block slides up a frictionless inclined plane.

From the above figure

    $x=\frac{h}{\sin \theta }$        (I)

Figure 2 represent the free body diagram of the block ignoring friction.

Write the expression for net force acting on the block in the $x$ direction.

    $\sum F_x=ma$

The above equation can be written as

    $\begin{array}{c}-mg\sin \theta =ma\\ a=-g\sin \theta \end{array}$        (II)

Write the expression for net force acting on the block in the $y$ direction.

    $\sum F_y=0$

The above equation can be written as

    $N-mg\cos \theta =0$

Write the kinematic equation.

    $v^2=v_0^2+2a\left(x-x_0\right)$

Here, $v$ is the final speed of the block, $v_0$ is the initial speed of the block, $a$ is the acceleration of the block, $x$ is the initial position of the block, and $x_0$ is the initial position of block.

Since the final velocity of the block is zero and the initial position of the block is zero, the above equation is written as

    $\begin{array}{c}0=v_0^2+2ax\\ v_0=\sqrt{-2ax}\end{array}$

Use equation (I) and (II) in the above equation.

    $\begin{array}{c}{v}_0=\sqrt{-2\left(-g\sin \theta \right)\frac{h}{\sin \theta }}\\ =\sqrt{2gh}\end{array}$        (III)

Conclusion:

Substitute $9.8\text{m}/{\text{s}}^2$ for $g$, and $4.5\text{m}$ for $h$ in equation (III), to find $v_0$.

    $\begin{array}{c}{v}_0=\sqrt{2\left(9.8\text{m}/{\text{s}}^2\right)\left(4.5\text{m}\right)}\\ =9.4\text{m}/\text{s}\end{array}$

Therefore, the initial speed of the block is $\underset{\_}{9.4\text{m}/\text{s}}$.

(b)

To determine

The initial speed required for the block to reach the maximum height.

Answer to Problem 83P

The initial speed required for the block to reach the maximum height is $\underset{\_}{11\text{m}/\text{s}}$.

Explanation of Solution

Figure 3 represent the free body diagram of the block by considering friction.

Write the expression for net force acting on the block in the $x$ direction.

    $\sum F_x=m{a}_x$

The above equation can be written as

    $-mg\sin \theta -F_t=m{a}_x$        (IV)

Write the expression for frictional force.

    $F_t={\mu }_{k}mg\cos \theta$

Here, $F_t$ is the frictional force, ${\mu }_k$ is the coefficient of kinetic friction, $m$ is the mass, and $g$ is the acceleration due to gravity.

Use the above equation in equation (IV).

    $\begin{array}{c}-mg\sin \theta -{\mu }_{k}mg\cos \theta =m{a}_x\\ a_x=-g\left(\sin \theta -{\mu }_k\cos \theta \right)\end{array}$        (V)

Write the kinematic equation.

    $v^2=v_0^2+2a\left(x-x_0\right)$

Here, $v$ is the final speed of the block, $v_0$ is the initial speed of the block, $a$ is the acceleration of the block, $x$ is the initial position of the block, and $x_0$ is the initial position of block.

Since the final velocity of the block is zero and the initial position of the block is zero, the above equation is written as

    $\begin{array}{c}0=v_0^2+2ax\\ v_0=\sqrt{-2ax}\end{array}$

Use $x=\frac{h}{\sin \theta }$ in the above equation.

    $v_0=\sqrt{\frac{-2ah}{\sin \theta }}$        (VI)

Conclusion:

Substitute $9.8\text{m}/{\text{s}}^2$ for $g$, $25°$ for $\theta$, and $0.15$ for ${\mu }_k$ in equation (V), to find $a_x$.

    $\begin{array}{c}{a}_x=-\left(9.8\text{m}/{\text{s}}^2\right)\left(\sin 25°-\left(0.15\right)\cos 25°\right)\\ =-5.5\text{m}/{\text{s}}^2\end{array}$

Substitute $-5.5\text{m}/\text{s}$ for $a$, $4.5\text{m}$ for $h$, and $25°$ for $\theta$ in equation (VI), to find $v_0$.

    $\begin{array}{c}{v}_0=\sqrt{\frac{-2\left(-5.5{\text{m}/\text{s}}^2\right)\left(4.5\text{ m}\right)}{\sin 25°}}\\ =11\text{m}/\text{s}\end{array}$

Therefore, the initial speed required for the block to reach the maximum height is $\underset{\_}{11\text{m}/\text{s}}$.