Newtons Laws EX 31

There are four forces acting on this block: $F_{G}$, $F_{N}$, $F_{f}$, $F$. The block slides up the incline and therefore the force of kinetic friction points down the incline (because friction opposes motion).

We select the x-axis parallel to the incline and draw a free body diagram:

The components of the forces exerted on the mass m = 5 kg are:

Forces	x-component	y-component
$F_{G}$	50 cos(-127°) = -30 N	50 sin(-127°) = -40 N
$F_{N}$	0	$F_{N}$
$F_{f}$	$-F_{f}$	0
$F$	25 cos(-37°) = 20 N	25 sin(-37°) = -15 N

Note that we used $g=10m/s^{2}$ in our computations.

First we have to deal with the y-component. Writing

$-40-15+F_{N}=0$

we see that $F_{N}=55N$. Substituting this value for $F_{N}$ in the equation for the force of kinetic friction we get:

$F_{f_{max}}=\mu _{K}F_{N}=(0.1)(55)=5.5N$

Now, looking at the x-component, we write

$-30+20-5.5=5a$

Solving for $a$, we find $a=-3.1m/s^{2}$.

To find the distance travelled in 2 s, we will use the equations of kinematics. The initial velocity is 6 m/s. We use the following equation for displacement:

$\Delta x=v_{i}t+{\frac {1}{2}}at^{2}$

Substituting, we find:

$\Delta x=(6)(2)+{\frac {1}{2}}(-3.1)(2)^{2}=5.8m$