Chapter 4, Problem 55P

To determine

(a) To Determine:

The acceleration of the block sliding down an inclined plane.

Answer to Problem 55P

Solution:

The acceleration of the block down the inclined plane is found to be 3.7 m/s².

Explanation of Solution

A block starting from rest slides down a distance s along the length of the incline which is at an angle $\theta$ and attains a speed v.

The free body diagram for the block is shown in the diagram below.

The weight mg of the block acts vertically downwards. The normal force $F_N$ acts perpendicular to the incline. The weight mg is resolved into two components, $mg\sin \theta \text{ and }mg\cos \theta$ parallel and perpendicular to the incline.

The block experiences a force due to the component of weight $mg\sin \theta$ and is accelerated at the rate a.

    $F=mg\sin \theta$……(1)

By Newton’s second law,

    $F=ma$……(2)

From equations (1)and (2),

    $\begin{array}{l}ma=mg\sin \theta \\ a=g\sin \theta \end{array}$……(3)

Given:

The initial speed of the block $v_0=0\text{ m/s}$

The angle of the incline $\theta =22°$

The distance traveled by the block down the incline $s=12.0\text{ m}$

Formula used:

    $a=g\sin \theta$$v^2=v_0^2+2as$

Calculation:

Substitute the given value of $\theta$ and 9.8 m/s² for g in the equation for acceleration

$\begin{array}{l}a=g\sin \theta \\ =\left(9.8{\text{ m/s}}^2\right)\left(\sin 22°\right)\\ =3.67{\text{ m/s}}^2\\ =3.7{\text{ m/s}}^2\end{array}$

To determine

(b) To Determine:

The speed of the block when it reaches the end of the incline.

Answer to Problem 55P

Solution:

The speed of the block when it reaches the end of the inclined plane is 9.4 m/s.

Explanation of Solution

To calculate the speed of the block when it reaches the end of the incline, use the equation of motion,

    $v^2=v_0^2+2as$

Given:

The initial speed of the block $v_0=0\text{ m/s}$

The angle of the incline $\theta =22°$

The distance traveled by the block down the incline $s=12.0\text{ m}$

Formula used:

    $v^2=v_0^2+2as$

Calculation:

Substitute the values of $v_0$, a and s in the equation $v^2=v_0^2+2as$ to calculate the speed of the block when it reaches the end of the inclined plane.

$\begin{array}{l}{v}^2=v_0^2+2as\\ =\left(0\text{ m/s}\right)+2\left(3.7{\text{ m/s}}^2\right)\left(12.0\text{ m}\right)\\ =88.8{\left(\text{m/s}\right)}^2\\ v=\sqrt{88.8{\left(\text{m/s}\right)}^2}\\ =9.42\text{ m/s}\\ =9.4\text{ m/s}\end{array}$

The speed of the body when reaches the end of the incline is calculated using the equation of motion and is found to be 9.4 m/s.