91Ó°ÊÓ

$$\begin{gathered} M=5.0kg,\mathrm{θ}=37° \\ F=50N,{\mathrm{μ}}_k=0.30 \end{gathered}$$

Frictional force is given by the product of coefficient of friction and the normal reaction. Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass. Using these concepts, the problem can be solved.

Formula:

$$\begin{gathered} {\mathbf{f}}_{\mathbf{k}}\mathbf{=}{\mathbf{μ}}_{\mathbf{k}}{\mathbf{F}}_{\mathbf{k}} \\ \mathbf{F}\mathbf{=}\mathbf{ma} \end{gathered}$$

(a)

Applying Newton’s second law to the x (directed uphill) and y (directed away from the incline surface) axes, we obtain

$$\begin{gathered} F\cos \mathrm{θ}-f_k-mg\sin \mathrm{θ}=ma \\ {F}_N-f\sin \mathrm{θ}-mg\cos \mathrm{θ}=ma \end{gathered}$$

Using. $f_k={\mathrm{μ}}_k{F}_N$

$$\begin{gathered} a=\frac{F}{m}\left(\cos \mathrm{θ}-{\mathrm{μ}}_k\sin \mathrm{θ}\right)-g\left(\sin \mathrm{θ}-{\mathrm{μ}}_k\cos \mathrm{θ}\right) \\ =\frac{50N}{5.0kg}\left(\cos 37°-0.37\sin 37°\right)-g\left(\cos 37°+0.37\sin 37°\right) \\ a=-2.1m/s^{2}or\left|a\right|=+2.1m/s^2 \end{gathered}$$

‘(²ú)

The assumed direction for acceleration in part (a) is up the inclined plane and after the calculation it came$a=-2.1m/s^2$ . This negative shows our assume direction is wrong. So, the direction of is down the plane.

(c)

$v_0=+4.0m/s$and v = 0,

$$\begin{gathered} ∆x=\frac{v^2-v_0^2}{2a} \\ =\frac{{\left(4m/s\right)}^2-0}{2\left(-2.1m/s^2\right)} \\ =3.9m \end{gathered}$$

(d)

we expect${\mathrm{μ}}_sâ‰\yen {\mathrm{μ}}_k$ ; otherwise, an object started into motion would immediately start decelerating (before it gained any speed)!

$$\begin{gathered} f_{s,max}={\mathrm{μ}}_s{F}_N \\ ={\mathrm{μ}}_S\left(F\sin \mathrm{θ}+mg\cos \mathrm{θ}\right) \end{gathered}$$

Which turns out to be 21 N. But acceleration along the x axis is zero, we must have

$$\begin{gathered} f_s=F\cos \mathrm{θ}-mg\sin \mathrm{θ} \\ =10N \end{gathered}$$

(The fact that this is positive reinforces our suspicion that$f_s$ points downhill).

Since the$f_s$ needed to remain at rest is less than$f_{s,max}$ then it stays at that location