91Ó°ÊÓ

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

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:

$F=ma$

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

$F_N=mg\cos \mathrm{θ}$

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

$$\begin{gathered} ma=mg\sin \mathrm{θ}-f_k \\ ma=mg\sin \mathrm{θ}-{\mathrm{μ}}_{k}mg\cos \mathrm{θ} \end{gathered}$$

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

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

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

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

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

$a=g\sin \mathrm{θ}+{\mathrm{μ}}_{k}g\cos \mathrm{θ}$

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

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

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

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