Question

1. -/4 points a block with a mass of 3.00 kg starts from rest at the top of a 29.5° incline and slides 2.00 m down the incline in 1.80 s. (a) what is the magnitude of the acceleration of the block (in m/s²)? m/s² (b) what is the frictional force (in n) acting on the block? (enter the magnitude.) n (c) what is the coefficient of kinetic friction between the block and the incline? (d) what is the speed of the block (in m/s) after it has slid 2.00 m? m/s

Explanation:

Step1: Find acceleration using kinematic - equation

The kinematic equation $x = v_0t+\frac{1}{2}at^{2}$, where $v_0 = 0$ (starts from rest), $x = 2.00\ m$, and $t = 1.80\ s$. Substituting $v_0 = 0$ into the equation gives $x=\frac{1}{2}at^{2}$. Solving for $a$:
$a=\frac{2x}{t^{2}}$
$a=\frac{2\times2.00}{1.80^{2}}\approx1.23\ m/s^{2}$

Step2: Analyze forces along the incline for frictional - force

The gravitational force component along the incline is $F_{g\parallel}=mg\sin\theta$ and the net - force along the incline is $F_{net}=ma$. According to Newton's second law $F_{net}=F_{g\parallel}-f$, where $m = 3.00\ kg$, $g = 9.8\ m/s^{2}$, $\theta=29.5^{\circ}$, and $a$ is the acceleration found in step 1.
$F_{g\parallel}=mg\sin\theta=3.00\times9.8\times\sin(29.5^{\circ})\approx3.00\times9.8\times0.492 = 14.47\ N$
$F_{net}=ma=3.00\times1.23 = 3.69\ N$
Since $F_{net}=F_{g\parallel}-f$, then $f=F_{g\parallel}-F_{net}$
$f = 14.47-3.69=10.78\ N$

Step3: Calculate coefficient of kinetic friction

The normal force on the block on the incline is $N = mg\cos\theta$.
$N=3.00\times9.8\times\cos(29.5^{\circ})\approx3.00\times9.8\times0.871 = 25.66\ N$
The frictional force $f=\mu_kN$, so $\mu_k=\frac{f}{N}$
$\mu_k=\frac{10.78}{25.66}\approx0.42$

Step4: Find the speed using kinematic - equation

Use the kinematic equation $v = v_0+at$, with $v_0 = 0$, $a = 1.23\ m/s^{2}$, and $t = 1.80\ s$
$v=0 + 1.23\times1.80=2.21\ m/s$

Answer:

(a) $1.23\ m/s^{2}$
(b) $10.78\ N$
(c) $0.42$
(d) $2.21\ m/s$