Question

In Fig. 6-54, the coefficient of kinetic friction between the block and inclined plane is 0.20, and angle $\mathit{\theta }$is $\mathbf{60}\mathbf{°}$. What are the

(a) magnitude aand

(b) direction (up or down the plane) of the block’s acceleration if the block is sliding down the plane?

What are (c) aand (d) the direction if the block is sent sliding up the plane?

Short answer

1. $a=7.5\mathrm{m}/{\mathrm{s}}^2$.
2. The direction of the acceleration$\overrightarrow{a}$ is down the slope.
3. $a=9.5\mathrm{m}/{\mathrm{s}}^2$.
4. The direction is down the slope.

Step-by-step solution

Given data

• The angle of inclination of the ramp, $\theta =60°$.
• The coefficient of kinetic friction, ${\mu }_k=0.2$.

To understand the concept

The problem deals with Newton’s second law of motion, which states that the acceleration a, of an object is dependent upon the net force F acting upon the object and the mass m, of the object when the system is in equilibrium.Use Newton’s law of motion to calculate the kinetic friction.

Formula:

$F=ma$

(a) Calculate the magnitude a (up or down the plane) of the block’s acceleration if the block is sliding down the plane.

The normal force on the block on the inclined ramp is:

$F_N=mg\cos \theta$

To calculate acceleration, use Newton’s second law of motion:

$$\begin{gathered} ma=mg\sin \theta -f_k \\ ma=mg\sin \theta -{\mu }_{k}mg\cos \theta \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} a=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times \sin \left(60°\right)-\left(0.2\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times \cos \left(60°\right) \\ a=7.5\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the magnitude of a is $7.5\mathrm{m}/{\mathrm{s}}^2$.

(b) Calculate the direction (up or down the plane) of the block’s acceleration if the block is sliding down the plane.

From the above calculation, we can conclude that the direction of the acceleration$\overrightarrow{a}$ is down the slope.

(c) Calculate the magnitude a (up or down the plane) of the block’s acceleration if the block is sliding up the plane.

When the friction force is the downhill direction, then the equation of motion will be,

$a=g\sin \theta +{\mu }_{k}g\cos \theta$

Substitute the values in the above expression, and we get,

$$\begin{gathered} a=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times \sin \left(60°\right)+\left(0.2\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times \cos \left(60°\right) \\ a=9.5\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the magnitude of a is $9.5\mathrm{m}/{\mathrm{s}}^2$.

(d) Calculate the direction of a (up or down the plane) of the block’s acceleration if the block is sliding up the plane.

From the above calculation, we can conclude that the direction is down the slope.