Question

a string passing over a pulley connects two blocks, one of mass 1.0 kg and one of mass 2.0 kg, as shown in the figure. the 1.0 kg block slides on a horizontal surface and the 2.0 kg block slides on an inclined surface 37° at from the horizontal. the coefficient of kinetic friction between each block and surface is 0.3. the masses of the string and pulley are negligible, and the pulley can rotate with negligible friction around its axle. as the block on the horizontal surface is sliding to the right as shown in the figure, the magnitude of the acceleration of the blocks is most nearly
(a) 1.4 m/s²
(b) 2.4 m/s²
(c) 3.0 m/s²
(d) 3.9 m/s²

Explanation:

Step1: Analyze forces on 1.0 - kg block

The normal force on the 1.0 - kg block $N_1=m_1g$, and the frictional force $f_1=\mu_kN_1=\mu_km_1g$. According to Newton's second - law, $T - f_1=m_1a$, so $T=\mu_km_1g + m_1a$.

Step2: Analyze forces on 2.0 - kg block

The normal force on the 2.0 - kg block $N_2 = m_2g\cos\theta$. The frictional force $f_2=\mu_kN_2=\mu_km_2g\cos\theta$. The component of the gravitational force along the incline is $m_2g\sin\theta$. According to Newton's second - law, $m_2g\sin\theta−T - f_2=m_2a$.

Step3: Substitute $T$ into the second - law equation of 2.0 - kg block

Substitute $T=\mu_km_1g + m_1a$ into $m_2g\sin\theta−T - f_2=m_2a$. We get $m_2g\sin\theta-(\mu_km_1g + m_1a)-\mu_km_2g\cos\theta=m_2a$.

Step4: Rearrange the equation to solve for $a$

$m_2g\sin\theta-\mu_km_1g-\mu_km_2g\cos\theta=(m_1 + m_2)a$. Then $a=\frac{m_2g\sin\theta-\mu_km_1g-\mu_km_2g\cos\theta}{m_1 + m_2}$.
Given $m_1 = 1.0\ kg$, $m_2 = 2.0\ kg$, $\mu_k = 0.3$, $\theta = 37^{\circ}$, and $g = 9.8\ m/s^2$. $\sin37^{\circ}\approx0.6$, $\cos37^{\circ}\approx0.8$.
$a=\frac{2\times9.8\times0.6-0.3\times1\times9.8-0.3\times2\times9.8\times0.8}{1 + 2}$
$=\frac{11.76 - 2.94-4.704}{3}=\frac{4.116}{3}\approx1.4\ m/s^2$.

Answer:

A. $1.4\ m/s^2$