Chapter 4, Problem 57P

The crate shown in Fig. 4-60 lies on a plane tilted at an angle $\theta =25.0°$ to the horizontal, with ${\mu }_k =\mathrm{0.19.}$(a) Determine the acceleration of the crate as it slides down the plane. (b) If the crate starts from rest 8.15 m up along the plane from its base, what will be the crate's speed when it reaches the bottom of the incline?

v

Figure 4-60

To determine

Part(a)To Determine:

The acceleration of the crate when it slides down an inclined plane when force of kinetic friction acts on it.

Answer to Problem 57P

Solution:

The acceleration of the crate down the inclined plane is found to be 2.5 m/s².

Explanation of Solution

The crate of mass slides down a plane inclined at an angle $\theta$ to the horizontal. The free body diagram of the crate is shown below.

The weight mg of the crate acts vertically downwards. The normal force $F_N$ acts perpendicular to the inclined plane. The weight mg is resolved into two components, $mg\sin \theta \text{ and }mg\cos \theta$ parallel and perpendicular to the inclined plane. The force of kinetic friction $F_{fr}$ acts opposite to the direction of motion of the box. There is a net force acting on the box in the downward direction parallel to the inclined plane accelerating it at a rate a.

Since there is no motion perpendicular to the incline,

    $F_N=mg\cos \theta$…… (1)

The force of kinetic friction is related to the normal force as,

    $F_{fr}\text{=}{\mu }_k{F}_N$…… (2)

Since the crate slides down the incline, there is a net force F that acts downwards. This is given by,

    $F=mg\sin \theta -F_{fr}$…… (3)

Substitute equations (1)and (2)in (3).

    $\begin{array}{c}F=mg\sin \theta -{\mu }_{k}mg\cos \theta \\ =mg\left(\sin \theta -{\mu }_k\cos \theta \right)\end{array}$……. (4)

From Newton’s second law,

    $F=ma$

Here, a is the resultant acceleration of the crate down the incline.

Therefore,

    $\begin{array}{c}ma=mg\left(\sin \theta -{\mu }_k\cos \theta \right)\\ a=g\left(\sin \theta -{\mu }_k\cos \theta \right)\end{array}$……(5)

Given:

The length of the incline $s=8.15\text{ m}$

The angle at which the incline is inclined $\theta =25.0°$

The coefficient of kinetic friction ${\mu }_k=0.19$

The initial velocity of the crate $v_0=0\text{ m/s}$

Formula used:

    $a=g\left(\sin \theta -{\mu }_k\cos \theta \right)$

Calculation:

Calculate the acceleration of the crate down the incline by substituting 9.8 m/s²for g and the given values for ${\mu }_k$ and $\theta$.

$\begin{array}{l}a=g\left(\sin \theta -{\mu }_k\cos \theta \right)\\ =\left(9.8{\text{ m/s}}^2\right)\left[\left(\sin 25.0°\right)-\left(0.19\right)\left(\cos 25.0°\right)\right]\\ =2.45{\text{ m/s}}^2\\ =2.5{\text{ m/s}}^2\end{array}$

To determine

Part(b)To Determine:

The speed of the crate when it reaches the bottom of the incline.

Answer to Problem 57P

Solution:

The speed of the crate at the bottom of the incline is found to be 6.3 m/s.

Explanation of Solution

To calculate the speed of the crate when it reaches the end of the incline of length s is calculated using the equation of motion,

    $v^2=v_0^2+2as$……(6)

Given:

The length of the incline $s=8.15\text{ m}$

The angle at which the incline is inclined $\theta =25.0°$

The coefficient of kinetic friction ${\mu }_k=0.19$

The initial velocity of the crate $v_0=0\text{ m/s}$

Formula used:

$v^2=v_0^2+2as$

Calculation:

Calculate the speed of the crate when it reaches the bottom of the incline by substituting the calculated value of a and the given values of s and $v_0$ in the equation,

$\begin{array}{l}{v}^2=v_0^2+2as\\ ={\left(0\text{ m/s}\right)}^2+2\left(2.45{\text{ m/s}}^2\right)\left(8.15\text{ m}\right)\\ =39.935{\left(\text{m/s}\right)}^2\\ v=6.32\text{ m/s}\end{array}$