3. a 2.00 - kg object a is connected with a light string across a pulley of negligible mass and friction to…

Explanation:

Step1: Analyze forces for problem 3(b)

For object A of mass $m_A = 2.00\ kg$ on the inclined - plane, the force along the plane is $F_{A}=m_A g\sin\theta$. For object B of mass $m_B = 3.00\ kg$, the force is $F_{B}=m_B g$. Let the acceleration of the system be $a$ and the tension in the string be $T$. Using Newton's second law $F = ma$, for object A: $T - m_A g\sin\theta=m_A a$, and for object B: $m_B g - T=m_B a$.

Step2: Solve the system of equations for problem 3(b)

Add the two equations: $(T - m_A g\sin\theta)+(m_B g - T)=(m_A + m_B)a$. This simplifies to $m_B g - m_A g\sin\theta=(m_A + m_B)a$. So, $a=\frac{m_B g - m_A g\sin\theta}{m_A + m_B}$. Substitute $m_A = 2.00\ kg$, $m_B = 3.00\ kg$, and $\theta = 40.0^{\circ}$: [ \begin{align*} a&=\frac{3.00g-2.00g\sin40.0^{\circ}}{2.00 + 3.00}\ &=\frac{3.00g-2.00g\times0.6428}{5.00}\ &=\frac{(3.00 - 1.2856)g}{5.00}\ &=\frac{1.7144g}{5.00}\ & = 0.343g \end{align*} ] To find the tension $T$, substitute $a$ into $T - m_A g\sin\theta=m_A a$. So, $T=m_A(g\sin\theta + a)$. Substitute $m_A = 2.00\ kg$, $g\sin\theta=0.6428g$, and $a = 0.343g$: [ \begin{align*} T&=2.00g(0.6428+0.343)\ &=2.00g\times0.9858\ & = 1.97g \end{align*} ]

Step3: Analyze forces for problem 4(a)

Let the acceleration of the system be $a$. For object A of mass $m_A = 1.00\ kg$, the force is $T_1=m_A a$. For object B of mass $m_B = 2.00\ kg$, $T_2 - T_1=m_B a$. For object C of mass $m_C = 4.00\ kg$, $m_C g - T_2=m_C a$.

Step4: Solve the system of equations for problem 4(a)

Add the three equations: $m_C g=(m_A + m_B + m_C)a$. So, $a=\frac{m_C g}{m_A + m_B + m_C}$. Substitute $m_A = 1.00\ kg$, $m_B = 2.00\ kg$, and $m_C = 4.00\ kg$: [ \begin{align*} a&=\frac{4.00g}{1.00 + 2.00+4.00}\ &=\frac{4.00g}{7.00}\ &=\frac{4}{7}g \end{align*} ]

Step5: Calculate tension in string 2 for problem 4(b)

From $m_C g - T_2=m_C a$, we can solve for $T_2$. Substitute $a=\frac{4}{7}g$ and $m_C = 4.00\ kg$: [ \begin{align*} T_2&=m_C g - m_C a\ &=4.00g-4.00g\times\frac{4}{7}\ &=4.00g\left(1 - \frac{4}{7}\right)\ &=\frac{12}{7}g \end{align*} ]

Step6: Calculate tension in string 1 for problem 4(c)

From $T_1=m_A a$, substitute $m_A = 1.00\ kg$ and $a=\frac{4}{7}g$: [ T_1 = 1.00g\times\frac{4}{7}=\frac{4}{7}g ]

Answer:

3(b) Acceleration $a = 0.343g$, Tension $T = 1.97g$ 4(a) Acceleration of each object $a=\frac{4}{7}g$ 4(b) Tension in string 2 $T_2=\frac{12}{7}g$ 4(c) Tension in string 1 $T_1=\frac{4}{7}g$