Question

a string is tied to a 3.2 kg object on a table and a 1.5 kg object hanging over a pulley (figure 12). the coefficient of kinetic friction between the 3.2 kg object and the table is 0.20
(a) calculate the acceleration of each object2.
(b) determine the magnitude of the tension in the string.
(c) how far will the objects move in 1.2 s if the initial velocity of the 3.2 kg object is 1.3 m/s right?

Explanation:

Step1: Define Variables and Forces

Let \( m_1 = 3.2 \, \text{kg} \) (object on table), \( m_2 = 1.5 \, \text{kg} \) (hanging object), \( \mu_k = 0.20 \), \( g = 9.8 \, \text{m/s}^2 \). For \( m_1 \): tension \( T \), friction \( f_k=\mu_k m_1 g \), acceleration \( a \) (right). For \( m_2 \): weight \( m_2 g \), tension \( T \), acceleration \( a \) (down).

Step2: Write Force Equations

For \( m_1 \): \( T - f_k = m_1 a \) → \( T - \mu_k m_1 g = m_1 a \).
For \( m_2 \): \( m_2 g - T = m_2 a \).

Step3: Solve for Acceleration (Part a)

Add the two equations: \( m_2 g - \mu_k m_1 g = (m_1 + m_2) a \).
Substitute values: \( (1.5)(9.8) - 0.20(3.2)(9.8) = (3.2 + 1.5) a \).
Calculate left side: \( 14.7 - 6.272 = 8.428 \).
Right side: \( 4.7 a \).
Thus, \( a = \frac{8.428}{4.7} \approx 1.79 \, \text{m/s}^2 \approx 1.8 \, \text{m/s}^2 \).

Step4: Solve for Tension (Part b)

Use \( m_2 g - T = m_2 a \) → \( T = m_2 (g - a) \).
Substitute: \( T = 1.5(9.8 - 1.79) = 1.5(8.01) \approx 12.0 \, \text{N} \).

Step5: Solve for Distance (Part c)

Use kinematic equation: \( d = v_0 t + \frac{1}{2} a t^2 \).
\( v_0 = 1.3 \, \text{m/s} \), \( t = 1.2 \, \text{s} \), \( a = 1.79 \, \text{m/s}^2 \).
Calculate: \( d = 1.3(1.2) + 0.5(1.79)(1.2)^2 \).
\( d = 1.56 + 0.5(1.79)(1.44) = 1.56 + 1.2888 \approx 2.85 \, \text{m} \approx 2.9 \, \text{m} \).

Answer:

s:
(a) Acceleration: \( \approx 1.8 \, \text{m/s}^2 \)
(b) Tension: \( \approx 12.0 \, \text{N} \)
(c) Distance: \( \approx 2.9 \, \text{m} \)