Question

The crate shown in Fig. 4–60 lies on a plane tilted at an angle $\theta =\mathbf{25}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$to the horizontal, with ${\mu }_{\mathrm{k}}=0.19$. (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?

FIGURE 4–60 Crate on inclined plane. Problems 59 and 60

Short answer

(a) The acceleration of the crate is $2.45\mathrm{m}/{\mathrm{s}}^2$.

(b) The crate’s speed is $6.32\mathrm{m}/\mathrm{s}$when it reaches the bottom of the incline.

Step-by-step solution

Step 1. Given data and assumptions

The friction force on the crate acts upward as the crate moves downward.

Given data:

The mass of the crate is $m$.

The angle of incline is $\theta =25.0°$.

The coefficient of kinetic friction is ${\mu }_{\mathrm{k}}=0.19$.

The initial speed of the crate along the x-axis is $v_{\mathrm{i}}=0$.

The displacement of the crate is $\Delta x=8.15\mathrm{m}$.

Let the final speed of the crate be $v_{\mathrm{f}}$when it reaches the bottom, and a be the acceleration of the crate.

Step 2. Calculation of the acceleration

Part (a)

The normal force acting on the crate is $N=mg\cos \theta$.

The kinetic friction force on the crate is:

$\begin{array}{c}{f}_{\mathrm{k}}={\mu }_{\mathrm{k}}N\\ ={\mu }_{\mathrm{k}}mg\cos \theta \end{array}$

The net force on the crate is:

$\begin{array}{c}F=mg\sin \theta -f_{\mathrm{k}}\\ =mg\sin \theta -{\mu }_{\mathrm{k}}mg\cos \theta \\ =mg\left(\sin \theta -{\mu }_{\mathrm{k}}\cos \theta \right)\end{array}$

Then, the acceleration of the crate along the x-axis is:

$\begin{array}{c}a=\frac{F}{m}\\ =\frac{mg\left(\sin \theta -{\mu }_{\mathrm{k}}\cos \theta \right)}{m}\\ =g\left(\sin \theta -{\mu }_{\mathrm{k}}\cos \theta \right)\end{array}$

Now, substituting the values in the above equation, you will get:

$\begin{array}{c}a=9.80\mathrm{m}/{\mathrm{s}}^2\times \left(\sin 25.0°-\left(0.19\times \cos 25.0°\right)\right)\\ =2.45\mathrm{m}/{\mathrm{s}}^2\end{array}$

Hence, the acceleration of the crate is $2.45\mathrm{m}/{\mathrm{s}}^2$.

Step 3. Calculation of the speed of the crate at the bottom

Part (b)

Now, you know

$\begin{array}{c}{v}_{\mathrm{f}}^2=v_{\mathrm{i}}^2+2a\Delta x\\ v_{\mathrm{f}}^2=0^2+2\times 2.45\mathrm{m}/{\mathrm{s}}^2\times 8.15\mathrm{m}\\ v_{\mathrm{f}}=6.32\mathrm{m}/\mathrm{s}\end{array}$

Hence, the crate’s speed is $6.32\mathrm{m}/\mathrm{s}$when it reaches the bottom of the incline.