**Example 4.13 Connected Objects****Goal** Use both the general method and the system approach to solve a connected two-body problem involving gravity and friction.**Problem** A block with mass \( m_1 = 3.90 \, \text{kg} \) and a ball with mass \( m_2 = 7.70 \, \text{kg} \) are connected by a light string that passes over a frictionless pulley, as shown in Figure 4.23a. The coefficient of kinetic friction between the block and the surface is 0.300.(a) Find the acceleration of the two objects and the tension in the string.(b) Check the answer for the acceleration by using the system approach.**Strategy** Connected objects are handled by applying Newton's second law separately to each object. The free-body diagrams for the block and the ball are shown in Figure 4.23b, with the \( +x \)-direction to the right and the \( +y \)-direction upwards. The magnitude of the acceleration for both objects has the same value, \( |a_1| = |a_2| = a \). The block with mass \( m_1 \) moves in the positive \( x \)-direction, and the ball with mass \( m_2 \) moves in the negative \( y \)-direction, so \( a_1 = -a_2 \). Using Newton's second law, we can develop two equations involving the unknowns \( T \) and \( a \) that can be solved simultaneously. In part (b), treat the two masses as a single object, with the gravity force on the ball increasing the combined object's speed and the friction force on the block retarding it. The tension forces then become internal and don’t appear in the second law.**Figures Explanation**- **Figure 4.23a**: Shows two objects connected by a light string that passes over a frictionless pulley. The block (\( m_1 \)) is on a surface, and the ball (\( m_2 \)) is hanging.- **Figure 4.23b**: Illustrates free-body diagrams for the objects. For \( m_1 \), the forces include tension \( T \), friction \( f_k \), and normal \( n \). For \( m_2 \), the forces include gravity \( m_2g \) and **Solution****(a)** Find the acceleration of the objects and the tension in the string.Write the components of Newton's second law for the cube of mass \( m_1 \):\[\Sigma F_x = T - f_k = m_1 a_1 \]\[\Sigma F_y = n - m_1 g = 0\]The equation for the y-component gives \( n = m_1 g \). Substitute this value for \( n \) and \( f_k = \mu_k n \) into the equation for the x-component:\[T - \mu_k m_1 g = m_1 a_1 \quad (1)\]Apply Newton's second law to the ball, recalling that \( a_2 = -a_1 \):\[\Sigma F_y = -m_2 g + T = m_2 a_2 = -m_2 a_1 \quad (2)\]Subtract Equation (2) from Equation (1), eliminating \( T \) and leaving an equation that can be solved for \( a_1 \) (substitution can also be used):\[m_2 g - \mu_k m_1 g = (m_1 + m_2) a_1\]\[a = \frac{m_2 g - \mu_k m_1 g}{m_1 + m_2}\]Substitute the given values to obtain the acceleration:\[a = \frac{(7.70 \, \text{kg}) (9.80 \, \text{m/s}^2) - (0.300) (3.90 \, \text{kg}) (9.80 \, \text{m/s}^2)}{3.90 \, \text{kg} + 7.70 \, \text{kg}}\]\[a = \quad \text{m/s}^2\]Substitute the value into Equation (1) to find the tension \( T \):\[T = \quad \text{N}\]**(b)** Find the acceleration using the system approach, where the system consists of two blocks.Apply Newton's second law to the system and solve for \( a \). (Use \( m_1, m_2, \mu_k,

Given: m1 = 3.9 kg m2 = 7.7 kg coefficient of kinetic friction, μk = 0.3

Problem A block with mass m, 7.70 kg are connected by a light string that passes over a frictionless pulley, as shown in Figure 4.23a. The coefficient of kinetic friction between the block and the surface is 0.300. 3.90 kg and a ball with mass m2 %3D %3D

Problem A block with mass m, 7.70 kg are connected by a light string that passes over a frictionless pulley, as shown in Figure 4.23a. The coefficient of kinetic friction between the block and the surface is 0.300. 3.90 kg and a ball with mass m2 %3D %3D

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

**Example 4.13 Connected Objects**

**Goal**
Use both the general method and the system approach to solve a connected two-body problem involving gravity and friction.

**Problem**
A block with mass \( m_1 = 3.90 \, \text{kg} \) and a ball with mass \( m_2 = 7.70 \, \text{kg} \) are connected by a light string that passes over a frictionless pulley, as shown in Figure 4.23a. The coefficient of kinetic friction between the block and the surface is 0.300.

(a) Find the acceleration of the two objects and the tension in the string.

(b) Check the answer for the acceleration by using the system approach.

**Strategy**
Connected objects are handled by applying Newton's second law separately to each object. The free-body diagrams for the block and the ball are shown in Figure 4.23b, with the \( +x \)-direction to the right and the \( +y \)-direction upwards. The magnitude of the acceleration for both objects has the same value, \( |a_1| = |a_2| = a \). The block with mass \( m_1 \) moves in the positive \( x \)-direction, and the ball with mass \( m_2 \) moves in the negative \( y \)-direction, so \( a_1 = -a_2 \). Using Newton's second law, we can develop two equations involving the unknowns \( T \) and \( a \) that can be solved simultaneously. In part (b), treat the two masses as a single object, with the gravity force on the ball increasing the combined object's speed and the friction force on the block retarding it. The tension forces then become internal and don’t appear in the second law.

**Figures Explanation**

- **Figure 4.23a**: Shows two objects connected by a light string that passes over a frictionless pulley. The block (\( m_1 \)) is on a surface, and the ball (\( m_2 \)) is hanging.

- **Figure 4.23b**: Illustrates free-body diagrams for the objects. For \( m_1 \), the forces include tension \( T \), friction \( f_k \), and normal \( n \). For \( m_2 \), the forces include gravity \( m_2g \) and

**Solution**

**(a)** Find the acceleration of the objects and the tension in the string.

Write the components of Newton's second law for the cube of mass \( m_1 \):

\[
\Sigma F_x = T - f_k = m_1 a_1
\]
\[
\Sigma F_y = n - m_1 g = 0
\]

The equation for the y-component gives \( n = m_1 g \). Substitute this value for \( n \) and \( f_k = \mu_k n \) into the equation for the x-component:

\[
T - \mu_k m_1 g = m_1 a_1 \quad (1)
\]

Apply Newton's second law to the ball, recalling that \( a_2 = -a_1 \):

\[
\Sigma F_y = -m_2 g + T = m_2 a_2 = -m_2 a_1 \quad (2)
\]

Subtract Equation (2) from Equation (1), eliminating \( T \) and leaving an equation that can be solved for \( a_1 \) (substitution can also be used):

\[
m_2 g - \mu_k m_1 g = (m_1 + m_2) a_1
\]

\[
a = \frac{m_2 g - \mu_k m_1 g}{m_1 + m_2}
\]

Substitute the given values to obtain the acceleration:

\[
a = \frac{(7.70 \, \text{kg}) (9.80 \, \text{m/s}^2) - (0.300) (3.90 \, \text{kg}) (9.80 \, \text{m/s}^2)}{3.90 \, \text{kg} + 7.70 \, \text{kg}}
\]

\[
a = \quad \text{m/s}^2
\]

Substitute the value into Equation (1) to find the tension \( T \):

\[
T = \quad \text{N}
\]

**(b)** Find the acceleration using the system approach, where the system consists of two blocks.

Apply Newton's second law to the system and solve for \( a \). (Use \( m_1, m_2, \mu_k,

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