Question

question 4.3a: a 67.0 - kg box is pulled by a rope making an angle of θ = 15.5° with the horizontal. the tension in the rope is 26.0 n. (assume the surface the box slides across is frictionless.) if the box starts from rest, determine the time required to move the box a horizontal distance of 2.55 m. s question 4.3b: the figure shows two blocks connected by a lightweight string across a frictionless pulley. the hanging block has mass m₁ = 5.80 kg and the second block has mass m₂ = 3.40 kg. the ramp is frictionless and inclined at an angle of θ = 36.0° from the horizontal. when released from rest, block 1 will descend and block 2 will move up the ramp. (assume the ramp is frictionless.) determine the magnitude of each blocks acceleration and the tension in the string. a = m/s² t = n

Explanation:

Question 4.3a

Step1: Find the horizontal - component of the tension

The horizontal - component of the tension is $T_x=T\cos\theta$, where $T = 26.0\ N$ and $\theta=15.5^{\circ}$. So $T_x = 26.0\cos(15.5^{\circ})\approx25.0\ N$.

Step2: Calculate the acceleration of the box

According to Newton's second law $F = ma$, where $F = T_x$ and $m = 67.0\ kg$. So the acceleration $a=\frac{T_x}{m}=\frac{25.0}{67.0}\approx0.373\ m/s^{2}$.

Step3: Use the kinematic equation

The kinematic equation for motion with constant acceleration is $x = v_0t+\frac{1}{2}at^{2}$. Since $v_0 = 0\ m/s$ and $x = 2.55\ m$, we have $x=\frac{1}{2}at^{2}$. Then $t=\sqrt{\frac{2x}{a}}=\sqrt{\frac{2\times2.55}{0.373}}\approx3.70\ s$.

Question 4.3b

Step1: Set up Newton's second - law equations for each block

For block 1 ($m_1$), taking downwards as positive, $F_{net1}=m_1g - T=m_1a$. For block 2 ($m_2$), along the incline, $F_{net2}=T - m_2g\sin\theta=m_2a$.

Step2: Solve the system of equations

From $m_1g - T=m_1a$ and $T - m_2g\sin\theta=m_2a$, we can add the two equations: $m_1g - m_2g\sin\theta=(m_1 + m_2)a$.
Substitute $m_1 = 5.80\ kg$, $m_2 = 3.40\ kg$, $\theta = 36.0^{\circ}$, and $g = 9.8\ m/s^{2}$ into the equation:
\[

$$\begin{align*} a&=\frac{m_1g - m_2g\sin\theta}{m_1 + m_2}\\ &=\frac{(5.80\times9.8)- (3.40\times9.8\times\sin(36.0^{\circ}))}{5.80 + 3.40}\\ &=\frac{(5.80\times9.8)- (3.40\times9.8\times0.5878)}{9.20}\\ &=\frac{56.84-(3.40\times5.7604)}{9.20}\\ &=\frac{56.84 - 19.5854}{9.20}\\ &=\frac{37.2546}{9.20}\approx4.05\ m/s^{2} \end{align*}$$

\]

Step3: Find the tension in the string

Substitute $a$ into the equation $T - m_2g\sin\theta=m_2a$. So $T=m_2(a + g\sin\theta)$.
\[

$$\begin{align*} T&=3.40\times(4.05+9.8\times0.5878)\\ &=3.40\times(4.05 + 5.7604)\\ &=3.40\times9.8104\\ &\approx33.4\ N \end{align*}$$

\]

Answer:

Question 4.3a: $3.70$
Question 4.3b: $a = 4.05\ m/s^{2}$, $T = 33.4\ N$