Question

In Fig. 6-62, a 5.0 kgblock is sent sliding up a plane inclined at $\mathit{\theta }\mathbf{=}\mathbf{37}\mathbf{°}$while a horizontal force of magnitude 50 Nacts on it. The coefficient of kinetic friction between block and plane is0.30. What are the (a) magnitude and (b) direction (up or down the plane) of the block’s acceleration? The block’s initial speed is 4.0 m/s. (c) How far up the plane does the block go? (d) When it reaches its highest point, does it remain at rest or slide back down the plane?

Short answer

(a)$\left|a\right|=+2.1m/s^2$

(b) Down the plane.

(c) 3.9 m

(d) Remains at rest

Step-by-step solution

Given

$$\begin{gathered} M=5.0kg,\theta =37° \\ F=50N,{\mu }_k=0.30 \end{gathered}$$

Understanding the concept

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:

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

Calculation for magnitude of the block’s acceleration

(a)

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

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

Using. $f_k={\mu }_k{F}_N$

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

Calculation for the direction of the block’s acceleration 

‘(b)

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.

Calculate how far up the plane does the block go

(c)

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

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

Find out if the block remains at rest or slides back down the plane when it reaches its highest point

(d)

we expect${\mu }_s\ge {\mu }_k$ ; otherwise, an object started into motion would immediately start decelerating (before it gained any speed)!

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

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

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

(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