Question

a 1.5 kg block is held at rest near the top of a rough incline that makes an angle of 37° from the horizontal. the block is then released and starts moving with a constant acceleration. when the block travels 2.0 m, its velocity is 4.0 m/s. the coefficient of kinetic friction between the surface of the incline and the block is most nearly
a 0.20
b 0.25
c 0.40
d 0.75

Explanation:

Step1: Find the acceleration

Use the kinematic - equation $v^{2}=v_{0}^{2}+2ax$. Given $v_{0} = 0\ m/s$, $v = 4.0\ m/s$, and $x = 2.0\ m$.
$a=\frac{v^{2}-v_{0}^{2}}{2x}=\frac{4^{2}-0^{2}}{2\times2}=\frac{16}{4}=4\ m/s^{2}$

Step2: Analyze the forces on the block

The forces along the incline are $mg\sin\theta$ (down - slope) and $F_f=\mu_kN$ (up - slope), where $N = mg\cos\theta$. According to Newton's second law $F_{net}=ma$, so $mg\sin\theta-\mu_kmg\cos\theta=ma$.
We can cancel out the mass $m$ from both sides of the equation: $g\sin\theta-\mu_kg\cos\theta=a$.

Step3: Solve for the coefficient of kinetic friction

Rearrange the equation $g\sin\theta-\mu_kg\cos\theta=a$ to solve for $\mu_k$.
$\mu_k=\frac{g\sin\theta - a}{g\cos\theta}$. Given $g = 9.8\ m/s^{2}$, $\theta = 37^{\circ}$, and $a = 4\ m/s^{2}$.
$\sin37^{\circ}\approx0.6$, $\cos37^{\circ}\approx0.8$.
$\mu_k=\frac{9.8\times0.6 - 4}{9.8\times0.8}=\frac{5.88 - 4}{7.84}=\frac{1.88}{7.84}\approx0.24\approx0.25$

Answer:

B. 0.25